Rs Aggarwal 2015 for Class 10 Math Chapter 8 - Trigonometric Ratios Of Complementary Angles

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Page No 339:

Question 1:

Without using trigonometric tables, evaluate:
(i) $\frac{\sin 16 °}{\cos 74 °}$
(ii) $\frac{\sec 11 °}{cosec 79 °}$
(iii) $\frac{\tan 27 °}{\cot 63 °}$
(iv) $\frac{\cos 35 °}{\sin 55 °}$
(v) $\frac{cosec 42 °}{\sec 48 °}$
(vi) $\frac{\cot 38 °}{\tan 52 °}$

Answer:

$\begin{array}{l} (i) \frac{s i n 16^{\circ}}{c o s 74^{^{\circ}}} \end{array} \begin{array}{l} = \frac{\sin (90^{\circ} - 74^{\circ})}{\cos 74^{\circ}} \end{array} \begin{array}{l} = \frac{\cos 74^{\circ}}{\cos 74^{\circ}} \end{array} [∵ \sin (90 - θ) = \cos θ] \begin{array}{l} = 1 \end{array} \begin{array}{l} (ii) \frac{\sec 11^{\circ}}{\cos ec 79^{\circ}} \end{array} \begin{array}{l} = \frac{\sec (90^{\circ} - 79^{\circ})}{\cos {ec79}^{\circ}} \end{array} \begin{array}{l} = \frac{\cos {ec79}^{\circ}}{\cos {ec79}^{\circ}} \end{array} [∵ s e c (90 - θ) = \cos e c θ] \begin{array}{l} = 1 \end{array} \begin{array}{l} (iii) \frac{\tan 27^{\circ}}{\cot 63^{\circ}} \end{array} \begin{array}{l} = \frac{\tan (90^{\circ} - 63^{\circ})}{\cot 63^{\circ}} \end{array} \begin{array}{l} = \frac{\cot 63^{\circ}}{\cot 63^{\circ}} [∵ \tan (90 - θ) = \cot θ] \end{array} \begin{array}{l} = 1 \end{array} \begin{array}{l} (iv) \frac{\cos 35^{\circ}}{\sin 55^{\circ}} \end{array} \begin{array}{l} = \frac{\cos (90^{\circ} - 55^{\circ})}{\sin 55^{\circ}} \end{array} \begin{array}{l} = \frac{\sin 55^{\circ}}{\sin 55^{\circ}} [∵ \sin (90 - θ) = \cos θ] \end{array} = 1 \begin{array}{l} (v) \frac{\cos {ec42}^{\circ}}{\sec 48^{\circ}} \end{array} \begin{array}{l} = \frac{\cos ec (90^{\circ} - 48^{\circ})}{\sec 48^{\circ}} \end{array} \begin{array}{l} = \frac{\sec 48^{\circ}}{\sec 48^{\circ}} [∵ s e c (90 - θ) = \cos e c θ] \end{array} \begin{array}{l} = 1 \end{array} \begin{array}{l} (vi) \frac{\cot 38^{\circ}}{\tan 52^{\circ}} \end{array} \begin{array}{l} = \frac{\cot (90^{\circ} - 52^{\circ})}{\tan 52^{\circ}} \end{array} \begin{array}{l} = \frac{\tan 52^{\circ}}{\tan 52^{\circ}} \end{array} [∵ \tan (90 - θ) = \cot θ] =1$

Page No 340:

Question 2:

Prove that:

(i) $sinθ \cos (90 ° - θ) + \sin (90 ° - θ) cosθ = 1$
(ii) $\frac{sinθ}{\cos (90 ° - θ)} + \frac{cosθ}{\sin (90 ° - θ)} = 2$
(iii) $\frac{sinθ \cos (90 ° - θ) cosθ}{\sin (90 ° - θ)} + \frac{cosθ \sin (90 ° - θ) sinθ}{\cos (90 ° - θ)}$
(iv) $\frac{\cos (90 ° - θ) \sec (90 ° - θ) tanθ}{cosec (90 ° - θ) \sin (90 ° - θ) \cot (90 ° - θ)} + \frac{\tan (90 ° - θ)}{cotθ} = 2$
(v) $\frac{\cos (90 ° - θ)}{1 + \sin (90 ° - θ)} + \frac{1 + \sin (90 ° - θ)}{\cos (90 ° - θ)} = 2 cosecθ$
(vi) $\frac{\sec (90 ° - θ) cosecθ - \tan (90 ° - θ) cotθ + \cos^{2} 25 ° + \cos^{2} 65 °}{3 \tan 27 ° \tan 63 °} = \frac{2}{3} [CBSE 2010]$
(vii) $cotθ \tan (90 ° - θ) - \sec (90 ° - θ) cosecθ + \sqrt{3} \tan 12 ° \tan 60 ° \tan 78 ° = 2 [CBSE 2010]$

Answer:

$\begin{array}{l} (i) LH S = \sin θ \cos (90^{0} - θ) + \sin (90^{0} - θ) \cos θ \end{array} \begin{array}{l} = \sin θ \sin θ + \cos θ \cos θ \end{array} \begin{array}{l} = \sin^{2} θ + \cos^{2} θ \end{array} \begin{array}{l} = 1 \end{array} \begin{array}{l} = RHS \end{array} \begin{array}{l} H e n c e pr o v e d . \end{array} \begin{array}{l} (ii) LHS= \frac{\sin θ}{\cos (90^{0} - θ)} + \frac{\cos θ}{\sin (90^{0} - θ)} \end{array} \begin{array}{l} = \frac{\sin θ}{\sin θ} + \frac{\cos θ}{\cos θ} \end{array} \begin{array}{l} = 1 + 1 \end{array} \begin{array}{l} = 2 \end{array} \begin{array}{l} = RHS \end{array} \begin{array}{l} Hence pr o v e d . \end{array} \begin{array}{l} (iii) LHS = \frac{\sin θ \cos (90^{0} - θ) \cos θ}{\sin (90^{0} - θ)} + \frac{\cos θ \sin (90^{0} - θ) \sin θ}{\cos (90^{0} - θ)} \end{array} \begin{array}{l} = \frac{\sin θ \sin θ \cos θ}{\cos θ} + \frac{\cos θ \cos θ \sin θ}{\sin θ} \end{array} \begin{array}{l} = \sin^{2} θ + \cos^{2} θ \end{array} \begin{array}{l} = 1 \end{array} \begin{array}{l} \begin{array}{l} = RHS \end{array} \begin{array}{l} H e n c e pr o v e d . \end{array} \end{array} \begin{array}{l} (iv) LHS = \frac{\cos (90^{0} - θ) \sec (90^{0} - θ) \tan θ}{\cos ec (90^{0} - θ) \sin (90^{0} - θ) \cot (90^{0} - θ)} + \frac{\tan (90^{0} - θ)}{\cot θ} \end{array} \begin{array}{l} = \frac{\sin θ \cos ec θ \tan θ}{\sec θ \cos θ \tan θ} + \frac{\cot θ}{\cot θ} \end{array} \begin{array}{l} = 1 + 1 \end{array} \begin{array}{l} = 2 \end{array} \begin{array}{l} = RHS \end{array} \begin{array}{l} H e n c e pr o v e d . \end{array} \begin{array}{l} (v) LHS = \frac{\cos (90^{0} - θ)}{1 + \sin (90^{0} - θ)} + \frac{1 + \sin (90^{0} - θ)}{\cos (90^{0} - θ)} \end{array} \begin{array}{l} = \frac{\sin θ}{1 + \cos θ} + \frac{1 + \cos θ}{\sin θ} \end{array} \begin{array}{l} = \frac{\sin^{2} θ + {(1 + \cos θ)}^{2}}{(1 + \cos θ) \sin θ} \end{array} \begin{array}{l} = \frac{\sin^{2} θ + 1 + \cos^{2} θ + 2 \cos θ}{(1 + \cos θ) \sin θ} \end{array} \begin{array}{l} = \frac{1 + 1 + 2 \cos θ}{(1 + \cos θ) \sin θ} \end{array} \begin{array}{l} = \frac{2 + 2 \cos θ}{(1 + \cos θ) \sin θ} \end{array} \begin{array}{l} = \frac{2 (1 + \cos θ)}{(1 + \cos θ) \sin θ} \end{array} \begin{array}{l} = 2 \frac{1}{\sin θ} \end{array} \begin{array}{l} = 2 \cos ec θ \end{array} \begin{array}{l} = RHS \end{array} \begin{array}{l} H e n c e pr o v e d . \end{array}$

$(vi) LHS = \frac{\sec (90 ° - θ) cosec θ - \tan (90 ° - θ) \cot θ + \cos^{2} 25 ° + \cos^{2} 65 °}{3 \tan 27 ° \tan 63 °} = \frac{cosec θ cosec θ - \cot θ \cot θ + \sin^{2} (90 ° - 25 °) + \cos^{2} 65 °}{3 \tan 27 ° \cot (90 ° - 63 °)} = \frac{{cosec}^{2} θ - \cot^{2} θ + \sin^{2} 65 ° + \cos^{2} 65 °}{3 \tan 27 ° \cot 27 °} = \frac{1 + 1}{3 \times \tan 27 ° \times \frac{1}{\tan 27 °}} = \frac{2}{3} = RHS$

$(vii) LHS = \cot θ \tan (90 ° - θ) - \sec (90 ° - θ) cosec θ + \sqrt{3} \tan 12 ° \tan 60 ° \tan 78 ° = \cot θ \cot θ - cosec θ cosec θ + \sqrt{3} \tan 12 ° \times \sqrt{3} \times \cot (90 ° - 78 °) = \cot^{2} θ - {cosec}^{2} θ + 3 \tan 12 ° \cot 12 ° = - 1 + 3 \times \tan 12 ° \times \frac{1}{\tan 12 °} = - 1 + 3 = 2 = RHS$

Page No 340:

Question 3:

Without using trigonometric tables, evaluate:
(i) $\frac{\sin 50 °}{\cos 40 °} + \frac{cosec 40 °}{\sec 50 °} - 4 \cos 52 ° cosec 38 °$
(ii) $\frac{\cos 75 °}{\sin 15 °} + \frac{\sin 12 °}{\cos 78 °} - \cos 18 ° cosec 72 °$
(iii) $\frac{2 cosec 67 °}{\sec 23 °} - \frac{\tan 70 °}{\cot 20 °} - \cos 0 ° + \tan 38 ° \tan 52 °$
(iv) $\frac{\tan 76 °}{\cot 14 °} + \frac{\sec 58 °}{cosec 32 °} - \sin 35 ° \sec 55 ° - 8 \sin^{2} 30 °$

Answer:

$\begin{array}{l} (i) \frac{s i n 50^{0}}{c o s 40^{0}} + \frac{c o s e c 40^{0}}{s e c 50^{0}} - 4 c o s 52^{0} c o s e c 38^{0} \end{array} \begin{array}{l} = \frac{\sin (90^{0} - 40^{0})}{\cos 40^{0}} + \frac{\cos e c (90^{0} - 50^{0})}{\sec 50^{0}} - 4 \cos (90^{0} - 38^{0}) \cos e c 38^{0} \end{array} \begin{array}{l} = \frac{\cos 40^{0}}{\cos 40^{0}} + \frac{\sec 50^{0}}{\sec 50^{0}} - 4 \sin 38^{0} \frac{1}{\sin 38^{0}} \end{array} \begin{array}{l} = 1 + 1 - 4 \end{array} \begin{array}{l} = - 2 \end{array} \begin{array}{l} (ii) \frac{\cos 75^{0}}{\sin 15^{0}} + \frac{\sin 12^{0}}{\cos 78^{0}} - \cos 18^{0} \cos {ec72}^{0} \end{array} \begin{array}{l} = \frac{\cos (90^{0} - 15^{0})}{\sin 15^{0}} + \frac{\sin (90^{0} - 78^{0})}{\cos 78^{0}} - \cos (90^{0} - 72^{0}) \cos {ec72}^{0} \end{array} \begin{array}{l} = \frac{\sin 15^{0}}{\sin 15^{0}} + \frac{\cos 78^{0}}{\cos 78^{0}} - \sin 72^{0} . \frac{1}{\sin 72^{0}} \end{array} \begin{array}{l} =1+1 - 1 \end{array} \begin{array}{l} = 1 \end{array} \begin{array}{l} (iii) \frac{2 \cos {ec67}^{0}}{\sec 23^{0}} - \frac{\tan 70^{0}}{\cot 20^{0}} - \cos 0^{0} + \tan 38^{0} \tan 52^{0} \end{array} \begin{array}{l} = \frac{2 \cos ec (90^{0} - 23^{0})}{\sec 23^{0}} - \frac{\tan (90^{0} - 20^{0})}{\cot 20^{0}} - 1 + \tan (90^{0} - 52^{0}) \tan 52^{0} \end{array} \begin{array}{l} = \frac{2 \sec 23^{0}}{\sec 23^{0}} - \frac{\cot 20^{0}}{\cot 20^{0}} - 1 + \cot 52^{0} . \frac{1}{\cot 52^{0}} \end{array} \begin{array}{l} =2 - 1 - 1 + 1 \end{array} \begin{array}{l} =1 \end{array} \begin{array}{l} (iv) \frac{\tan 76^{0}}{\cot 14^{0}} + \frac{\sec 58^{0}}{\cos {ec32}^{0}} - \sin 35^{0} \sec 55^{0} - 8 \sin^{2} 30^{0} \end{array} \begin{array}{l} = \frac{\tan (90^{0} - 14^{0})}{\cot 14^{0}} + \frac{\sec (90^{0} - 32^{0})}{\cos {ec32}^{0}} - \sin (90^{0} - 55^{0}) \sec 55^{0} - 8 \times {(\frac{1}{2})}^{2} \end{array} \begin{array}{l} = \frac{\cot 14^{0}}{\cot 14^{0}} + \frac{\cos {ec32}^{0}}{\cos {ec32}^{0}} - \cos 55^{0} . \frac{1}{\cos 55^{0}} - 8 \times \frac{1}{4} \end{array} =1+1 - 1 - 2 = - 1$

Page No 340:

Question 4:

Without using trigonometric tables, prove that:
(i) cos 81° − sin 9° = 0
(ii) tan 71° − cot 19° = 0
(iii) cosec 80° − sec 10° = 0
(iv) cosec²72° − tan²18° = 1
(v) cos²75° + cos²15° = 1
(vi) tan²66° − cot²24° = 0
(vii) sin²48° + sin²42° = 1
(viii) cos²57° − sin²33° = 0
(ix) (sin 65° + cos 25°)(sin 65° − cos 25°) = 0

Answer:

$\begin{array}{l} (i) {LHS=cos81}^{0} - \sin 9^{0} \end{array} \begin{array}{l} =cos (90^{0} - 9^{0}) - \sin 9^{0} \end{array} \begin{array}{l} {=sin9}^{0} - \sin 9^{0} \end{array} \begin{array}{l} =0 \end{array} \begin{array}{l} =RHS \end{array} \begin{array}{l} (ii) {LHS=tan71}^{0} - \cot 19^{0} \end{array} \begin{array}{l} =tan (90^{0} - 19^{0}) - \cot 19^{0} \end{array} \begin{array}{l} {=cot19}^{0} - \cot 19^{0} \end{array} \begin{array}{l} =0 \end{array} \begin{array}{l} =RHS \\ \begin{array}{l} (iii) {LHS=cosec80}^{0} - \sec 10^{0} \end{array} \end{array} \begin{array}{l} =cosec (90^{0} - 10^{0}) - \sec 10^{0} \end{array} \begin{array}{l} {=sec10}^{0} - \sec 10^{0} \end{array} \begin{array}{l} =0 \end{array} \begin{array}{l} =RHS \end{array} \begin{array}{l} (iv) {LHS=cosec}^{2} 72^{0} - \tan^{2} 18^{0} \end{array} \begin{array}{l} {=cosec}^{2} (90^{0} - 18^{0}) - \tan^{2} 18^{0} \end{array} \begin{array}{l} {=sec}^{2} 18^{0} - \tan^{2} 18^{0} \end{array} \begin{array}{l} =1 \end{array} \begin{array}{l} =RHS \end{array} \begin{array}{l} (v) {LHS=cos}^{2} 75^{0} + \cos^{2} 15^{0} \end{array} \begin{array}{l} {=cos}^{2} (90^{0} - 15^{0}) + \cos^{2} 15^{0} \end{array} \begin{array}{l} {=sin}^{2} 15^{0} + \cos^{2} 15^{0} \end{array} =1 . =RHS \begin{array}{l} (vi) {LHS=tan}^{2} 66^{0} - \cot^{2} 24^{0} \end{array} \begin{array}{l} {=tan}^{2} (90^{0} - 24^{0}) - \cot^{2} 24^{0} \end{array} \begin{array}{l} {=cot}^{2} 24^{0} - \cot^{2} 24^{0} \end{array} \begin{array}{l} =0 \end{array} \begin{array}{l} =RHS \end{array} \begin{array}{l} (vii) {LHS=sin}^{2} 48^{0} + \sin^{2} 42^{0} \end{array} \begin{array}{l} {=sin}^{2} (90^{0} - 42^{0}) + \sin^{2} 42^{0} \end{array} \begin{array}{l} {=cos}^{2} 42^{0} + \sin^{2} 42^{0} \end{array} \begin{array}{l} =1 \end{array} \begin{array}{l} =RHS \end{array} \begin{array}{l} (viii) {LHS=cos}^{2} 57^{0} - \sin^{2} 33^{0} \end{array} \begin{array}{l} {=cos}^{2} (90^{0} - 33^{0}) - \sin^{2} 33^{0} \end{array} \begin{array}{l} {=sin}^{2} 33^{0} - \sin^{2} 33^{0} \end{array} \begin{array}{l} =0 \end{array} \begin{array}{l} =RHS \end{array} \begin{array}{l} (ix) LHS= ({sin65}^{0} + \cos 25^{0}) ({sin65}^{0} - \cos 25^{0}) \end{array} \begin{array}{l} {=sin}^{2} 65^{0} - \cos^{2} 25^{0} \end{array} \begin{array}{l} {=sin}^{2} (90^{0} - 25^{0}) - \cos^{2} 25^{0} \end{array} \begin{array}{l} {=cos}^{2} 25^{0} - \cos 25^{0} \end{array} \begin{array}{l} =0 \end{array} =RHS$

Page No 340:

Question 5:

Without using trigonometric tables, prove that:

(i) sin53° cos37° + cos53° sin37° = 1
(ii) cos54° cos36° − sin54° sin36° = 0
(iii) sec70° sin20° + cos20° cosec70° = 2
(iv) sin35° sin55° − cos35° cos55° = 0
(v) (sin72° + cos18°)(sin72° − cos18°) = 0
(vi) tan48° tan23° tan42° tan67° = 1

Answer:

$\begin{array}{l} (i) LHS = \sin 53^{0} \cos 37^{0} + \cos 53^{0} \sin 37^{0} \end{array} \begin{array}{l} = \sin (90^{0} - 37^{0}) \cos 37^{0} + \cos (90^{0} - 37^{0}) \sin 37^{0} \end{array} \begin{array}{l} = \cos 37^{0} \cos 37^{0} + \sin 37^{0} \sin 37^{0} \end{array} \begin{array}{l} = \cos^{2} 37^{0} + \sin^{2} 37^{0} \end{array} \begin{array}{l} = 1 \end{array} \begin{array}{l} = RHS \end{array} \begin{array}{l} (ii) LHS = \cos 54^{0} \cos 36^{0} - \sin 54^{0} \sin 36^{0} \end{array} \begin{array}{l} = \cos (90^{0} - 36^{0}) \cos 36^{0} - \sin (90^{0} - 36^{0}) \sin 36^{0} \end{array} \begin{array}{l} = \sin 36^{0} \cos 36^{0} - \cos 36^{0} \sin 36^{0} \end{array} \begin{array}{l} = 0 \end{array} \begin{array}{l} = RHS \end{array} \begin{array}{l} (iii) LHS = \sec 70^{0} \sin 20^{0} + \cos 20^{0} cosec 70^{0} \end{array} \begin{array}{l} = \sec (90^{0} - 20^{0}) \sin 20^{0} + \cos 20^{0} cosec (90^{0} - 20^{0}) \end{array} \begin{array}{l} = cosec 20^{0} . \frac{1}{cosec 20^{0}} + \frac{1}{\sec 20^{0}} . \sec 20^{0} \end{array} \begin{array}{l} = 1 + 1 \end{array} \begin{array}{l} = 2 \end{array} = RHS$

$(iv) LHS = \sin 35 ° \sin 55 ° - \cos 35 ° \cos 55 ° = \sin 35 ° \cos (90 ° - 55 °) - \cos 35 ° \sin (90 - 55 °) = \sin 35 ° \cos 35 ° - \cos 35 ° \sin 35 ° = 0 = RHS$

$(v) LHS = (\sin 72 ° + \cos 18 °) (\sin 72 ° - \cos 18 °) = (\sin 72 ° + \cos 18 °) [\cos (90 ° - 72 °) - \cos 18 °] = (\sin 72 ° + \cos 18 °) (\cos 18 ° - \cos 18 °) = (\sin 72 ° + \cos 18 °) (0) = 0 = RHS$

$(vi) LHS = \tan 48 ° \tan 23 ° \tan 42 ° \tan 67 ° = \cot (90 ° - 48 °) \cot (90 ° - 23 °) \tan 42 ° \tan 67 ° = \cot 42 ° \cot 67 ° \tan 42 ° \tan 67 ° = \frac{1}{\tan 42 °} \times \frac{1}{\tan 67 °} \times \tan 42 ° \times \tan 67 ° = 1 = RHS$

Page No 340:

Question 6:

Prove that:

(i) tan5° tan25° tan30° tan65° tan85° = $\frac{1}{\sqrt{3}}$
(ii) cot12° cot38° cot52° cot60° cot78° = $\frac{1}{\sqrt{3}}$
(iii) cos15° cos35° cosec55° cos60° cosec75° = $\frac{1}{2}$
(iv) cos1° cos2° cos3° ... cos180° = 0
(v) ${(\frac{\sin 49 °}{\cos 41 °})}^{2} + {(\frac{\cos 41 °}{\sin 49 °})}^{2} = 2$

Answer:

$\begin{array}{l} (i) {LHS=tan5}^{0} \tan 25^{0} \tan 30^{0} \tan 65^{0} \tan 85^{0} \end{array} \begin{array}{l} =tan (90^{0} - 85^{0}) \tan (90^{0} - 65^{0}) × \frac{1}{\sqrt{3}} × \frac{1}{\cot 65^{0}} \frac{1}{\cot 85^{0}} \end{array} \begin{array}{l} {=cot85}^{0} \cot 65^{0} \frac{1}{\sqrt{3}} \frac{1}{\cot 65^{0}} \frac{1}{\cot 85^{0}} \end{array} = \frac{1}{\sqrt{3}} = RHS$

$(ii) LHS = \cot 12 ° \cot 38 ° \cot 52 ° \cot 60 ° \cot 78 ° = \tan (90 ° - 12 °) \times \tan (90 ° - 38 °) \times \cot 52 ° \times \frac{1}{\sqrt{3}} \times \cot 78 ° = \frac{1}{\sqrt{3}} \times \tan 78 ° \times \tan 52 ° \times \cot 52 ° \times \cot 78 ° = \frac{1}{\sqrt{3}} \times \tan 78 ° \times \tan 52 ° \times \frac{1}{\tan 52 °} \times \frac{1}{\tan 78 °} = \frac{1}{\sqrt{3}} = RHS$

$\begin{array}{l} (iii) {LHS=cos15}^{0} \cos 35^{0} \cos {ec55}^{0} \cos 60^{0} \cos {ec75}^{0} \end{array} \begin{array}{l} =cos (90^{0} - 75^{0}) \cos (90^{0} - 55^{0}) \frac{1}{\sin 55^{0}} × \frac{1}{2} × \frac{1}{\sin 75^{0}} \end{array} \begin{array}{l} {=sin75}^{0} \sin 55^{0} \frac{1}{\sin 55^{0}} \times \frac{1}{2} \times \frac{1}{\sin 75^{0}} \end{array} = \frac{1}{2} = RHS$

$(iv) LHS = \cos 1 ° \cos 2 ° \cos 3 ° . . . \cos 180 ° = \cos 1 ° \times \cos 2 ° \times \cos 3 ° \times . . . \times \cos 90 ° \times . . . \cos 180 ° = \cos 1 ° \times \cos 2 ° \times \cos 3 ° \times . . . \times 0 \times . . . \cos 180 ° = 0 = RHS$

$(v) LHS = {(\frac{\sin 49 °}{\cos 41 °})}^{2} + {(\frac{\cos 41 °}{\sin 49 °})}^{2} = {(\frac{\cos (90 ° - 49 °)}{\cos 41 °})}^{2} + {(\frac{\cos 41 °}{\cos (90 ° - 49 °)})}^{2} = {(\frac{\cos 41 °}{\cos 41 °})}^{2} + {(\frac{\cos 41 °}{\cos 41 °})}^{2} = 1^{2} + 1^{2} = 1 + 1 = 2 = RHS$

Disclaimer: The RHS of (v) given in textbook is incorrect. There should be 2 instead 1. The same has been corrected in the solution here.

Page No 341:

Question 7:

Prove that
(i) sin (70° + θ) − cos (20° − θ) = 0
(ii) tan (55° − θ) − cot (35° + θ) = 0
(iii) cosec (67° + θ) − sec (23° − θ) = 0
(iv) cosec (65 °+ θ) sec (25° − θ) − tan (55° − θ) + cot (35° + θ) = 0
(v) sin (50° + θ ) − cos (40° − θ) + tan 1° tan 10° tan 80° tan 89° = 1.

Answer:

$\begin{array}{l} (i) L .H .S=sin (70^{0} + θ) - \cos (20^{0} - θ) \end{array} \begin{array}{l} =sin {90^{0} - (20^{0} - θ)} - \cos (20^{0} - θ) \end{array} \begin{array}{l} = \cos (20^{0} - θ) - \cos (20^{0} - θ) \end{array} \begin{array}{l} =0 \\ \begin{array}{l} =R .H .S . \end{array} \end{array} \begin{array}{l} (ii) L .H .S=tan (55^{0} - θ) - \cot (35^{0} + θ) \end{array} \begin{array}{l} =tan {90^{0} - (35^{0} + θ)} - \cot (35^{0} + θ) \end{array} \begin{array}{l} = \cot (35^{0} + θ) - \cot (35^{0} + θ) \end{array} \begin{array}{l} =0=R .H .S . \end{array} \begin{array}{l} (iii) L .H .S=cosec (67^{0} + θ) - \sec (23^{0} - θ) \end{array} \begin{array}{l} =cosec {90^{0} - (23^{0} - θ)} - \sec (23^{0} - θ) \end{array} \begin{array}{l} = \sec (23^{0} - θ) - \sec (23^{0} - θ) \end{array} =0=R .H .S . \begin{array}{l} (iv) L .H .S=cosec (65^{0} + θ) - \sec (25^{0} - θ) - \tan (55^{0} - θ) + \cot (35^{0} + θ) \end{array} \begin{array}{l} = \cos e c {90^{0} - (25^{0} - θ)} - \sec (25^{0} - θ) - \tan (55^{0} - θ) + \cot {90^{0} - (55^{0} - θ)} \end{array} \begin{array}{l} = \sec (25^{0} - θ) - \sec (25^{0} - θ) - \tan (55^{0} - θ) + \tan (55^{0} - θ) \end{array} \begin{array}{l} = 0 = R .H .S \end{array} \begin{array}{l} (v) L .H .S = \sin (50^{0} + θ) - \cos (40^{0} - θ) + \tan 1^{0} \tan 10^{0} \tan 80^{0} \tan 89^{0} \end{array} \begin{array}{l} = \sin {90^{0} - (40^{0} - θ)} - \cos (40^{0} - θ) + {\tan 1^{0} \tan (90^{0} - 1^{0})} {\tan 10^{0} \tan (90^{0} - 10^{})} \end{array} \begin{array}{l} = \cos (40^{0} - θ) - \cos (40^{0} - θ) + (\tan 1^{0} \cot 1^{0}) (\tan 10^{0} \cot 10^{0}) \end{array} \begin{array}{l} = (\frac{1}{\cot 1^{0}} \times \cot 1^{0}) (\tan 10^{0} \times \frac{1}{\tan 10^{0}}) \end{array} \begin{array}{l} = 1 \times 1 \end{array} = 1 = R .H .S$

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Question 8:

Express each of the following in terms of trigonometric ratios of angles lying between 0° and 45°.

(i) sin67° + cos75°
(ii) cot65° + tan49°
(iii) sec78° + cosec56°
(iv) cosec54° + sin72°

Answer:

$(i) \sin 67 ° + \cos 75 ° = \cos (90 ° - 67 °) + \sin (90 ° - 75 °) = \cos 23 ° + \sin 15 °$

$(ii) \cot 65 ° + \tan 49 ° = \tan (90 ° - 65 °) + \cot (90 ° - 49 °) = \tan 25 ° + \cot 41 °$

$(iii) \sec 78 ° + cosec 56 ° = \sec (90 ° - 12 °) + cosec (90 ° - 34 °) = cosec 12 ° + \sec 34 °$

$(iv) cosec 54 ° + \sin 72 ° = \sec (90 ° - 54 °) + \cos (90 ° - 72 °) = \sec 36 ° + \cos 18 °$

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Question 9:

If sin 3 A = cos (A − 26°), where 3 A is an acute angle, find the value of A.

Answer:

$\begin{array}{l} sin3A = \cos (A - 26^{0}) \end{array} \begin{array}{l} ⇒cos (90^{0} - 3 A) = \cos (A - 26^{0}) [∵ \sin θ = \cos (90^{0} - θ)] \end{array} \begin{array}{l} \Rightarrow 90^{0} - 3 A = A - 26^{0} \end{array} \begin{array}{l} \Rightarrow 116^{0} = 4 A \end{array} \Rightarrow A = \frac{116^{0}}{4} = 29^{0}$

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Question 10:

If tan 2 A = cot (A − 12°), where 2 A is an acute angle, find the value of A.

Answer:

$\begin{array}{l} tan2 A = \cot (A - 12^{0}) \end{array} \begin{array}{l} =>cot (90^{0} - 2 A) = \cot (A - 12^{0}) [∵ \tan θ = \cot (90^{0} - θ)] \end{array} \begin{array}{l} => (90^{0} - 2 A) = (A - 12^{0}) \end{array} \begin{array}{l} {=>102}^{0} = 3 A \end{array} => A = \frac{102^{0}}{3} = 34^{0}$

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Question 11:

If sec 4 A = cosec (A − 15°), where 4 A is an acute angle, find the value of A.

Answer:

$\begin{array}{l} sec 4 A = \cos e c (A - 15^{0}) \end{array} \begin{array}{l} => cosec (90^{0} - 4 A) = \cos e c (A - 15^{0}) [∵ \sec θ = \cos e c (90^{0} - θ)] \end{array} \begin{array}{l} {=>90}^{0} - 4 A = A - 15^{0} \end{array} \begin{array}{l} {=>105}^{0} = 5 A \end{array} => A = \frac{105^{0}}{5} = 21^{0}$

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Question 12:

If A, B, C are the angles of a ΔABC, prove that tan ( $\frac{B + C}{2}$ ) = cot $\frac{A}{2}$ .

Answer:

$\begin{array}{l} {We know that the sum of the angles of a triangle is 180}^{0} . \\ \begin{array}{l} ∴ A + B + C = 180^{0} \end{array} \end{array} \begin{array}{l} => B + C = 180^{0} - A \end{array} \begin{array}{l} => \frac{B + C}{2} = 90^{0} - \frac{A}{2} [D ividing both sides by 2] \end{array} \begin{array}{l} => \tan (\frac{B + C}{2}) = \tan (90^{0} - \frac{A}{2}) \end{array} \begin{array}{l} => \tan (\frac{B + C}{2}) = \cot \frac{A}{2} [∵ \tan (90^{0} - θ) = \cot θ] \end{array} Hence proved.$

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Question 13:

$\frac{\sin 15 ° \cos 75 ° + \cos 15 ° \sin 75 °}{\tan 5 ° \tan 30 ° \tan 35 ° \tan 55 ° \tan 85 °}$

Answer:

$\begin{array}{l} \frac{\sin 15^{0} \cos 75^{0} + \cos 15^{0} \sin 75^{0}}{\tan 5^{0} \tan 30^{0} \tan 35^{0} \tan 55^{0} \tan 85^{0}} \end{array} \begin{array}{l} = \frac{\sin 15^{0} \cos (90^{0} - 15^{0}) + \cos 15^{0} \sin (90^{0} - 15^{0})}{\tan 30^{0} {\tan 5^{0} \tan 35^{0} \tan 55^{0} \tan 85^{0}}} \end{array} \begin{array}{l} = \frac{\sin 15^{0} \sin 15^{0} + \cos 15^{0} \cos 15^{0}}{\frac{1}{\sqrt{3}} {\tan (90^{0} - 5^{0}) \tan 85^{0} . \tan 35^{0} \tan (90^{0} - 35^{0})}} \end{array} \begin{array}{l} = \frac{\sin^{2} 15^{0} + \cos^{2} 15^{0}}{\frac{1}{\sqrt{3}} (\cot 85^{0} \tan 85^{0}) (\tan 35^{0} . \cot 35^{0})} \end{array} \begin{array}{l} = \frac{\sqrt{3}}{(\frac{1}{\tan 85^{0}} \tan 85^{0}) (\frac{1}{\cot 35^{0}} . \cot 35^{0})} \end{array} \begin{array}{l} = \frac{\sqrt{3}}{1 \times 1} \end{array} = \sqrt{3}$

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Question 14:

$\frac{3 \cos 55 °}{7 \sin 35 °} - \frac{4 (\cos 70 ° \cos e c 20 °)}{7 (\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °)}$

Answer:

$\begin{array}{l} \frac{3 \cos 55^{0}}{7 \sin 35^{0}} - \frac{4 (\cos 70^{0} \cos ec 20^{0})}{7 (\tan 5^{0} \tan 25^{0} \tan 45^{0} \tan 65^{0} \tan 85^{0})} \end{array} \begin{array}{l} = \frac{3 \cos (90^{0} - 35^{0})}{7 \sin 35^{0}} - \frac{4 (\cos 70^{0} \frac{1}{\sin 20^{0}})}{7 \times 1 {\tan 5^{0} \tan (90^{0} - 5^{0}) . \tan 25^{0} \tan (90^{0} - 65^{0})}} \end{array} \begin{array}{l} = \frac{3 \sin 35}{7 \sin 35^{0}} - \frac{4 (\cos 70^{0} \frac{1}{\sin (90^{0} - 70^{0})})}{7 (\tan 5^{0} \cot 5^{0}) (\tan 25^{0} \cot 25^{0})} \end{array} \begin{array}{l} = \frac{3}{7} - \frac{4 (\cos 70^{0} \frac{1}{\cos 70^{0}})}{7 (\tan 5^{0} \frac{1}{\tan 5^{0}}) (\tan 25^{0} \frac{1}{\tan 25^{0}})} \end{array} \begin{array}{l} = \frac{3}{7} - \frac{4}{7} \end{array} = \frac{3 - 4}{7} = \frac{- 1}{7}$

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Question 15:

Prove that:

$\frac{2}{3} {cosec}^{2} 58 ° - \frac{2}{3} \cot 58 ° \tan 32 ° - \frac{5}{3} \tan 13 ° \tan 37 ° \tan 45 ° \tan 53 ° \tan 77 ° = - 1$

Answer:

$\begin{array}{l} \frac{2}{3} \cos {ec}^{2} 58^{0} - \frac{2}{3} \cot 58^{0} \tan 32^{0} - \frac{5}{3} \tan 13^{0} \tan 37^{0} \tan 45^{0} \tan 53^{0} \tan 77^{0} \end{array} \begin{array}{l} = \frac{2}{3} (\cos {ec}^{2} 58^{0} - \cot 58^{0} \tan 32^{0}) - \frac{5}{3} \tan 13^{0} \tan (90^{0} - 13^{0}) \tan 37^{0} \tan (90^{0} - 37^{0}) (\tan 45^{0}) \end{array} \begin{array}{l} = \frac{2}{3} {\cos {ec}^{2} 58^{0} - \cot 58^{0} \tan (90^{0} - 58^{0})} - \frac{5}{3} \tan 13^{0} \cot 13^{0} \tan 37^{0} \cot 37^{0} (1) \end{array} \begin{array}{l} = \frac{2}{3} (\cos {ec}^{2} 58^{0} - \cot 58^{0} \cot 58^{0}) - \frac{5}{3} \tan 13^{0} \frac{1}{\tan 13^{0}} \tan 37^{0} \frac{1}{\tan 37^{0}} \end{array} \begin{array}{l} = \frac{2}{3} (\cos {ec}^{2} 58^{0} - \cot^{2} 58^{0}) - \frac{5}{3} \end{array} \begin{array}{l} = \frac{2}{3} - \frac{5}{3} \end{array} = - 1$

Hence Proved

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Question 16:

$\frac{\cos 70 °}{\sin 20 °} + \frac{\cos 55 ° cosec 35 °}{\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °}$

Answer:

$\begin{array}{l} \frac{\cos 70^{0}}{\sin 20^{0}} + \frac{\cos 55^{0} \cos ec 35^{0}}{\tan 5^{0} \tan 25^{0} \tan 45^{0} \tan 65^{0} \tan 85^{0}} \end{array} \begin{array}{l} = \frac{\cos (90^{0} - 20^{0})}{\sin 20^{0}} + \frac{\cos 55^{0} \cos ec (90^{0} - 55^{0})}{\tan 45^{0} \tan 25^{0} \tan (90^{0} - 25^{0}) \tan (90^{0} - 5^{0}) \tan 5^{0}} \end{array} \begin{array}{l} = \frac{\sin 20^{0}}{\sin 20^{0}} + \frac{\cos 55^{0} \sec 55^{0}}{1 \times \tan 25^{0} \cot 25^{0} \cot 5^{0} \tan 5^{0}} \end{array} \begin{array}{l} = 1 + \frac{\cos 55^{0} \frac{1}{\cos 55^{0}}}{\tan 25^{0} \frac{1}{\tan 25^{0}} \cot 5^{0} \frac{1}{\cot 5^{0}}} \end{array} =1+ \frac{1}{1} = 1 + 1 = 2$

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Question 17:

$\frac{\sin 70 °}{\cos 20 °} + \frac{cosec 36 °}{\sec 54 °} - \frac{2 \cos 43 ° cosec 47 °}{\tan 10 ° \tan 40 ° \tan 50 ° \tan 80 °}$

Answer:

$\begin{array}{l} \frac{\sin 70^{0}}{\cos 20^{0}} + \frac{\cos ec 36^{0}}{\sec 54^{0}} - \frac{2 \cos 43^{0} \cos ec 47^{0}}{\tan 10^{0} \tan 40^{0} \tan 50^{0} \tan 80^{0}} \end{array} \begin{array}{l} = \frac{\sin (90^{0} - 20^{0})}{\cos 20^{0}} + \frac{\cos ec (90^{0} - 54^{0})}{\sec 54^{0}} - \frac{2 \cos 43^{0} \cos ec (90^{0} - 43^{0})}{\tan 10^{0} \tan (90^{0} - 10^{0}) \tan 40^{0} \tan (90^{0} - 40^{0})} \end{array} \begin{array}{l} = \frac{\cos 20^{0}}{\cos 20^{0}} + \frac{\sec 54^{0}}{\sec 54^{0}} - \frac{2 \cos 43^{0} \sec 43^{0}}{\tan 10^{0} \cot 10^{0} \tan 40^{0} \cot 40^{0}} \end{array} \begin{array}{l} = 1 + 1 - \frac{2 \cos 43^{0} \frac{1}{\cos 43^{0}}}{\tan 10^{0} \frac{1}{\tan 10^{0}} \tan 40^{0} \frac{1}{\tan 40^{0}}} \end{array} \begin{array}{l} = 2 - 2 \end{array} =0$

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Question 18:

$\frac{- tanθ \cdot \cot (90 ° - θ) + secθ cosec (90 ° - θ) + \sin^{2} 35 ° + \sin^{2} 55 °}{\tan 10 ° \tan 20 ° \tan 30 ° \tan 70 ° \tan 80 °}$

Answer:

$\begin{array}{l} \frac{- \tan θ \cot (90^{0} - θ) + \sec θ \cos ec (90^{0} - θ) + \sin^{2} 35^{0} + \sin^{2} 55^{0}}{\tan 10^{0} \tan 20^{0} \tan 30^{0} \tan 70^{0} \tan 80^{0}} \end{array} \begin{array}{l} = \frac{- \tan θ \tan θ + \sec θ \sec θ + \sin^{2} 35^{0} + \sin^{2} (90^{0} - 35^{0})}{{\tan 10^{0} \tan (90^{0} - 10^{0}) \tan 20^{0} \tan (90^{0} - 20^{0})} \tan 30^{0}} \end{array} \begin{array}{l} = \frac{- \tan^{2} θ + \sec^{2} θ + (\sin^{2} 35^{0} + \cos^{2} 35^{0})}{(\tan 10^{0} \cot 10^{0}) (\tan 20^{0} \cot 20^{0}) \frac{1}{\sqrt{3}}} \end{array} \begin{array}{l} = \frac{\sec^{2} θ - \tan^{2} θ + 1}{(\tan 10^{0} \frac{1}{\tan 10^{0}}) (\tan 20^{0} \frac{1}{\tan 20^{0}}) \frac{1}{\sqrt[]{3}}} \end{array} \begin{array}{l} = \frac{(1 + 1) \sqrt{3}}{1} \end{array} = 2 \sqrt{3}$

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Question 19:

tan 7° tan 23° tan 60° tan 67° tan 83° + $\frac{\cot 54 °}{\tan 36 °}$ + sin 20° sec 70° − 2

Answer:

$\begin{array}{l} {tan7}^{0} \tan 23^{0} \tan 60^{0} \tan 67^{0} \tan 83^{0} + \frac{\cot 54^{0}}{\tan 36^{0}} + \sin 20^{0} \sec 70^{0} - 2 \end{array} \begin{array}{l} = (\tan 60^{0}) (\tan 7^{0} \tan 83^{0}) (\tan 23^{0} \tan 67^{0}) + \frac{\cot (90^{0} - 36^{0})}{\tan 36^{0}} + \sin 20^{0} \sec (90^{0} - 20^{0}) - 2 \end{array} \begin{array}{l} = \sqrt{3} {\tan 7^{0} \tan (90^{0} - 7^{0})} {\tan 23^{0} \tan (90^{0} - 23^{0})} + \frac{\tan 36^{0}}{\tan 36^{0}} + \sin 20^{0} \cos ec 20^{0} - 2 \end{array} \begin{array}{l} = \sqrt{3} (\tan 7^{0} \cot 7^{0}) (\tan 23^{0} \cot 23^{0}) + 1 + \sin 20^{0} \frac{1}{\sin 20^{0}} - 2 \end{array} \begin{array}{l} = \sqrt{3} \times (1 \times 1) + 1 + 1 - 2 \end{array} \begin{array}{l} = \sqrt{3} \times 1 \end{array} = \sqrt{3}$

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Question 20:

(sin²25° + sin²65°) + $\sqrt{3}$ (tan 5° tan 15° tan 30° tan 75° tan 85°)

Answer:

$\begin{array}{l} (\sin^{2} 25^{0} + \sin^{2} 65^{0}) + \sqrt{3} (\tan 5^{0} \tan 15^{0} \tan 30^{0} \tan 75^{0} \tan 85^{0}) \\ \begin{array}{l} =^{2} 25^{0} + \sin^{2} (90^{0} - 25^{0}) + \sqrt{3} (\tan 30^{0}) (\tan 5^{0} \tan 85^{0}) (\tan 15^{0} \tan 75^{0}) \end{array} \end{array} \begin{array}{l} = (\sin^{2} 25^{0} + \cos^{2} 25^{0}) + \sqrt{3} \times \frac{1}{\sqrt{3}} {\tan 5^{0} \tan (90^{0} - 5^{0})} {\tan 15^{0} \tan (90^{0} - 15^{0})} \end{array} \begin{array}{l} = 1 + 1 (\tan 5^{0} \cot 5^{0}) (\tan 15^{0} \cot 15^{0}) \end{array} \begin{array}{l} {=1+tan5}^{0} \frac{1}{\tan 5^{0}} \tan 15^{0} \frac{1}{\tan 15^{0}} \end{array} \begin{array}{l} =1+1 \end{array} =2$

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Question 21:

$\frac{\sin^{2} 40 ° + \sin^{2} 50 °}{\cos^{2} 20 ° + \cos^{2} 70 °} + \tan 10 ° \tan 20 ° \tan 60 ° \tan 70 ° \tan 80 °$

Answer:

$\begin{array}{l} \frac{\sin^{2} 40^{0} + \sin^{2} 50^{0}}{\cos^{2} 20^{0} + \cos^{2} 70^{0}} + \tan 10^{0} \tan 20^{0} \tan 60^{0} \tan 70^{0} \tan 80^{0} \end{array} \begin{array}{l} = \frac{\sin^{2} 40^{0} + \sin^{2} (90^{0} - 40^{0})}{\cos^{2} 20^{0} + \cos^{2} (90^{0} - 20^{0})} + (\tan 60^{0}) \tan 20^{0} \tan (90^{0} - 20^{0}) \tan 10^{0} \tan (90^{0} - 10^{0}) \end{array} \begin{array}{l} = \frac{\sin^{2} 40^{0} + \cos^{2} 40^{0}}{\cos^{2} 20^{0} + \sin^{2} 20^{0}} + \sqrt{3} (\tan 20^{0} \cot 20^{0} \tan 10^{0} \cot 10^{0}) \end{array} \begin{array}{l} =1+ \sqrt{3} (\tan 20^{0} \frac{1}{\tan 20^{0}} \tan 10^{0} \frac{1}{\tan 10^{0}}) \end{array} = (1+ \sqrt{3})$

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Question 22:

$\frac{\sec^{2} 54 ° - \cot^{2} 36 °}{{cosec}^{2} 57 ° - \tan^{2} 33 °} + 2 \sin^{2} 38 ° \sec^{2} 52 ° - \sin^{2} 45 °$

Answer:

$\begin{array}{l} \frac{\sec^{2} 54^{0} - \cot^{2} 36^{0}}{\cos {ec}^{2} 57^{0} - \tan^{2} 33^{0}} + 2 \sin^{2} 38^{0} \sec^{2} 52^{0} - \sin^{2} 45^{0} \end{array} \begin{array}{l} = \frac{\sec^{2} 54^{0} - \cot^{2} (90^{0} - 54^{0})}{\cos {ec}^{2} 57^{0} - \tan^{2} (90^{0} - 57^{0})} + 2 \sin^{2} 38^{0} \sec^{2} (90^{0} - 38^{0}) - {(\frac{1}{\sqrt{2}})}^{2} \end{array} \begin{array}{l} = \frac{\sec^{2} 54^{0} - \tan^{2} 54^{0}}{\cos {ec}^{2} 57^{0} - \cot^{2} 57^{0}} + 2 \sin^{2} 38^{0} \cos {ec}^{2} 38^{0} - \frac{1}{2} \end{array} \begin{array}{l} {=1+2sin}^{2} 38^{0} \frac{1}{\sin^{2} 38^{0}} - \frac{1}{2} \end{array} \begin{array}{l} =1+2 - \frac{1}{2} \end{array} \begin{array}{l} =3 - \frac{1}{2} \end{array} = \frac{5}{2}$

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Question 23:

$\frac{\sec^{2} (90 ° - θ) - \cot^{2} θ}{2 (\sin^{2} 25 ° + \sin^{2} 65 °)} + \frac{2 \cos^{2} 60 ° \tan^{2} 28 ° \tan^{2} 62 °}{3 (\sec^{2} 43 ° - \cot^{2} 47 °)}$

Answer:

$\begin{array}{l} \frac{\sec^{2} (90^{0} - θ) - \cot^{2} θ}{2 (\sin^{2} 25^{0} + \sin^{2} 65^{0})} + \frac{2 \cos^{2} 60^{0} \tan^{2} 28^{0} \tan^{2} 62^{}}{3 (\sec^{2} 43^{0} - \cot^{2} 47^{0})} \end{array} \begin{array}{l} = \frac{\cos {ec}^{2} θ - \cot^{2} θ}{2 {\sin^{2} 25^{0} + \sin^{2} (90^{0} - 25^{0})}} + \frac{2 {(\frac{1}{2})}^{2} \tan^{2} 28^{0} \tan^{2} (90^{0} - 28^{0})}{3 {\sec^{2} 43^{0} - \cot^{2} (90^{0} - 43^{0})}} \end{array} \begin{array}{l} = \frac{1}{2 (\sin^{2} 25^{0} + \cos^{2} 25^{0})} + \frac{2 \times \frac{1}{4} \tan^{2} 28^{0} \cot^{2} 28^{0}}{3 (\sec^{2} 43^{0} - \tan^{2} 43^{0})} \end{array} \begin{array}{l} = \frac{1}{2} + \frac{\frac{1}{2} \times \tan^{2} 28^{0} \frac{1}{\tan^{2} 28^{0}}}{3} \end{array} \begin{array}{l} = \frac{1}{2} + \frac{1}{6} \end{array} \begin{array}{l} = \frac{3 + 1}{6} \end{array} = \frac{4}{6} = \frac{2}{3}$

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Question 24:

$\frac{\sec^{2} θ - \cot^{2} (90 ° - θ)}{{cosec}^{2} 67 ° - \tan^{2} 23 °} + (\sin^{2} 40 ° + \sin^{2} 50 °)$

Answer:

$\begin{array}{l} \frac{\sec^{2} θ - \cot^{2} (90^{0} - θ)}{\cos {ec}^{2} 67^{0} - \tan^{2} 23^{0}} + (\sin^{2} 40^{0} + \sin^{2} 50^{0}) \end{array} \begin{array}{l} = \frac{\sec^{2} θ - \tan^{2} θ}{\cos {ec}^{2} 67^{0} - \tan^{2} (90^{0} - 67^{0})} + \sin^{2} 40^{0} + \sin^{2} (90^{0} - 40^{0}) \end{array} \begin{array}{l} = \frac{1}{\cos {ec}^{2} 67^{0} - \cot^{2} 67^{0}} + (\sin^{2} 40^{0} + \cos^{2} 40^{0}) \end{array} \begin{array}{l} =1+1 \end{array} =2$

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Question 25:

$\frac{\sec^{2} 54 ° - \cot^{2} 36 °}{{cosec}^{2} 57 ° - \tan^{2} 33 °} + 2 \sin^{2} 38 ° \sec^{2} 52 ° - \sin^{2} 45 °$

Answer:

$\begin{array}{l} \frac{\sec^{2} 54^{0} - \cot^{2} 36^{0}}{\cos {ec}^{2} 57^{0} - \tan^{2} 33^{0}} + 2 \sin^{2} 38^{0} \sec^{2} 52^{0} - \sin^{2} 45^{0} \end{array} \begin{array}{l} = \frac{\sec^{2} 54^{0} - \cot^{2} (90^{0} - 54^{0})}{\cos {ec}^{2} 57^{0} - \tan^{2} (90^{0} - 57^{0})} + 2 \sin^{2} 38^{0} \sec^{2} (90^{0} - 38^{0}) - \sin^{2} 45^{0} \end{array} \begin{array}{l} = \frac{\sec^{2} 54^{0} - \tan^{2} 54^{0}}{\cos {ec}^{2} 57^{0} - \cot^{2} 57^{0}} + 2 \sin^{2} 38^{0} \cos {ec}^{2} 38^{0} - {(\frac{1}{\sqrt{2}})}^{2} \end{array} \begin{array}{l} {=1+2sin}^{2} 38^{0} \frac{1}{\sin^{2} 38^{0}} - \frac{1}{2} \end{array} \begin{array}{l} =1+2 - \frac{1}{2} \end{array} \begin{array}{l} =3 - \frac{1}{2} \end{array} = \frac{5}{2}$

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Question 26:

$\frac{3 \tan 25 ° \tan 40 ° \tan 50 ° \tan 65 ° - \frac{1}{2} \tan^{2} 60 °}{4 (\cos^{2} 29 ° + \cos^{2} 61 °)}$

Answer:

$\begin{array}{l} \frac{3 \tan 25^{0} \tan 40^{0} \tan 50^{0} \tan 65^{0} - \frac{1}{2} \tan^{2} 60^{0}}{4 (\cos^{2} 29^{0} + \cos^{2} 61^{0})} \end{array} \begin{array}{l} = \frac{3 \tan 25^{0} \tan (90^{0} - 25^{0}) \tan 40^{0} \tan (90^{0} - 40^{0}) - \frac{1}{2} {(\sqrt{3})}^{2}}{4 {\cos^{2} 29 + \cos^{2} (90^{0} - 29^{0})}} \end{array} \begin{array}{l} = \frac{3 \tan 25^{0} \cot 25^{0} \tan 40^{0} \cot 40^{0} - \frac{3}{2}}{4 (\cos^{2} 29^{0} + \sin^{2} 29^{0})} \end{array} \begin{array}{l} = \frac{3 \tan 25^{0} \frac{1}{\tan 25^{0}} \tan 40^{0} \frac{1}{\tan 40^{0}} - \frac{3}{2}}{4} \end{array} \begin{array}{l} = \frac{3 - \frac{3}{2}}{4} \end{array} \begin{array}{l} = \frac{\frac{6 - 3}{2}}{4} \end{array} = \frac{6 - 3}{8} = \frac{3}{8}$

Page No 342:

Question 27:

$\frac{2}{3} {cosec}^{2} 58 ° - \frac{2}{3} \cot 58 ° \tan 32 ° - \frac{5}{3} \tan 13 ° \tan 37 ° \tan 45 ° \tan 53 ° \tan 77 °$

Answer:

$\begin{array}{l} \frac{2}{3} \cos {ec}^{2} 58^{0} - \frac{2}{3} \cot 58^{0} \tan 32^{0} - \frac{5}{3} \tan 13^{0} \tan 37^{0} \tan 45^{0} \tan 53^{0} \tan 77^{0} \end{array} \begin{array}{l} = \frac{2}{3} (\cos {ec}^{2} 58^{0} - \cot 58^{0} \tan 32^{0}) - \frac{5}{3} \tan 13^{0} \tan (90^{0} - 13^{0}) \tan 37^{0} \tan (90^{0} - 37^{0}) (\tan 45^{0}) \end{array} \begin{array}{l} = \frac{2}{3} {\cos {ec}^{2} 58^{0} - \cot 58^{0} \tan (90^{0} - 58^{0})} - \frac{5}{3} \tan 13^{0} \cot 13^{0} \tan 37^{0} \cot 37^{0} (1) \end{array} \begin{array}{l} = \frac{2}{3} (\cos {ec}^{2} 58^{0} - \cot 58^{0} \cot 58^{0}) - \frac{5}{3} \tan 13^{0} \frac{1}{\tan 13^{0}} \tan 37^{0} \frac{1}{\tan 37^{0}} \end{array} \begin{array}{l} = \frac{2}{3} (\cos {ec}^{2} 58^{0} - \cot^{2} 58^{0}) - \frac{5}{3} \end{array} \begin{array}{l} = \frac{2}{3} - \frac{5}{3} \end{array} = - 1$

Page No 343:

Question 1:

$\frac{\tan 30 °}{\cot 60 °} = ?$
(a) $\frac{1}{\sqrt{2}}$
(b) $\frac{1}{\sqrt{3}}$
(c) $\sqrt{3}$
(d) 1

Answer:

(d) 1

$\frac{t a n 30 °}{c o t 60 °} = t a n 30 ° t a n 60 ° [∵ \tan θ = \frac{1}{c o t θ}] = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1$

Page No 343:

Question 2:

$\frac{\tan 35 °}{\cot 55 °} + \frac{\cot 78 °}{\tan 12 °} = ?$
(a) 0
(b) 1
(c) 2
(d) None of these

Answer:

$(c) 2 We have: \frac{\tan 35^{0}}{\cot 55^{0}} + \frac{\cot 78^{0}}{\tan 12^{o}} = \frac{\tan 35^{0}}{\cot (90^{0} - 35^{0})} + \frac{\cot (90^{0} - 12^{0})}{\tan 12^{0}} = \frac{\tan 35^{0}}{\tan 35^{0}} + \frac{\tan 12^{0}}{\tan 12^{0}} [∵ \cot (90^{0} - θ) = \tan θ] = 1 + 1 = 2$

Page No 343:

Question 3:

${(\frac{\tan 25 °}{cosec 65 °})}^{2} + {(\frac{\cot 25 °}{\sec 25 °})}^{2} = ?$
(a) 1
(b) 2
(c) $\frac{3}{4}$
(d) 8

Answer:

$(a) 1 We have: {(\frac{\tan 25^{0}}{\cos ec6 5^{0}})}^{2} + {(\frac{\cot 25^{0}}{\sec 65^{o}})}^{2} = {[\frac{\tan (90^{0} - 65^{0})}{\cos ec6 5^{0}}]}^{2} + {[\frac{\cot (90^{0} - 65^{0})}{\sec 65^{0}}]}^{2} [∵ \tan (90^{0} - θ) = \cot θ and cot (90^{0} - θ) = \tan θ] = {(\frac{\cos 65^{0}}{\sin 65^{0}} \times \sin 65^{0})}^{2} + {(\frac{\sin 65^{0}}{\cos 65^{0}} \times \cos 65^{0})}^{2} = \cos^{2} 65^{0} + \sin^{2} 65^{0} = 1$

Page No 343:

Question 4:

cos 1° cos 2° cos 3° ... cos 180° = ?
(a) −1
(b) 1
(c) 0
(d) $\frac{1}{2}$

Answer:

$(c) 0 \cos 1^{0} \cos 2^{0} \cos 3^{0} . . . \cos 180^{0} = \cos 1^{0} \cos 2^{0} \cos 3^{0} . . . \cos 90^{0} . . . \cos (180)^{0} = 0 [∵ \cos 90 ° = 0]$

Page No 343:

Question 5:

tan 10° tan 15° tan 75° tan 80° = ?
(a) $\sqrt{3}$
(b) $\frac{1}{\sqrt{3}}$
(c) −1
(d) 1

Answer:

$(d) 1 We have: {tan10}^{0} \tan 15^{0} \tan 75^{0} \tan 80^{0} = \tan 10^{0} \times \tan 15^{0} \times \tan (90^{0} - 15^{0}) \times \tan (90^{0} - 10^{0}) = \tan 10^{0} \times \tan 15^{0} \times \cot 15^{0} \times \cot 10^{0} [∵ \tan (90^{0} - θ) = \cot θ] = 1$

Page No 343:

Question 6:

tan 1° tan 2° tan 3° ... tan 89° = ?
(a) 1
(b) 0
(c) Cannot be determined
(d) None of these

Answer:

$(a) 1 We have: {tan1}^{0} \tan 2^{0} \tan 3^{0} . . . \tan 89^{0} = \tan 1^{0} \tan 2^{0} . . . \tan 45^{0} . . . \tan 88^{0} \tan 89^{0} = \tan 1^{0} \tan 2^{0} ... \tan 45^{0} ... \tan 88^{0} \tan 89^{0} = \tan 1^{0} \tan 2^{0} ... t a n 45^{0} ... t an (90^{0} - 2^{0}) \tan (90^{0} - 1^{0}) = \tan 1^{0} \tan 2^{0} . . . \cot 2^{0} \cot 1^{0} [∵ \tan (90^{0} - θ) = \cot θ {and tan45}^{0} = 1] = \frac{1}{\cot 1^{0}} \times \frac{1}{\cot 2^{0}} . . . 1 . . . \cot 2^{0} \times \cot 1^{0} = 1$

Page No 343:

Question 7:

tan 5° tan 25° tan 30° tan 65° tan 85° = ?
(a) $\sqrt{3}$
(b) $\frac{1}{\sqrt{3}}$
(c) 1
(d) none of these

Answer:

$The correct option is (b) . We have: {tan5}^{0} \tan 25^{0} \tan 30^{0} . \tan 65^{0} t a n 85^{0} = {tan5}^{0} \tan 25^{0} \tan 30^{0} \tan (90^{0} - 25^{0}) t a n (90^{0} - 5^{0}) = \tan 5^{0} \tan 25^{0} \times \frac{1}{\sqrt{3}} \times \cot 25^{0} \cot 5^{0} [∵ \tan (90^{0} - θ) = \cot θ {and tan30}^{0} = \frac{1}{\sqrt{3}}] = \frac{1}{\sqrt{3}}$

Page No 343:

Question 8:

cot 15° cot 16° cot 17° ... cot 73° cot 74° cot 75° = ?
(a) $\frac{1}{2}$
(b) 0
(c) 1
(d) −1

Answer:

$(c) 1 We have: {cot15}^{0} \cot 16^{0} \cot 17^{0} .. . \cot 73^{0} \cot 74^{0} \cot 75^{0} = {cot15}^{0} . \cot 16^{0} . \cot 17^{0} ... \cot 45^{0} ... \cot 73^{0} . \cot 74^{0} . \cot 75^{0} = {cot15}^{0} . \cot 16^{0} . \cot 17^{0} ... \cot 45^{0} ... \cot (90^{0} - 17^{0}) . \cot (90^{0} - 16^{0}) . \cot (90^{0} - 15^{0}) = {cot15}^{0} . \cot 16^{0} . \cot 17 ...1 . . . \tan 17^{0} . \tan 16^{0} . \tan 15^{0} [∵ \cot (90^{0} - θ) = \tan θ {and cot45}^{0} = 1] = 1$

Page No 343:

Question 9:

$\frac{2 \sin^{2} 63 ° + 1 + 2 \sin^{2} 27 °}{3 \cos^{2} 17 ° - 2 + 3 \cos^{2} 73 °} = ?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 2
(d) 3

Answer:

$(d) 3 Given: \frac{2 \sin^{2} 63^{0} + 1 + 2 \sin^{2} 27^{0}}{3 \cos^{2} 17^{0} - 2 + 3 \cos^{2} 73^{0}} = \frac{2 (\sin^{2} 63^{0} + \sin^{2} 27^{0}) + 1}{3 (\cos^{2} 17^{0} + \cos^{2} 73^{0}) - 2} = \frac{2 [\sin^{2} 63^{0} + \sin^{2} (90^{0} - 63^{0})] + 1}{3 [\cos^{2} 17^{0} + \cos^{2} (90^{0} - 17^{0})] - 2} = \frac{2 (\sin^{2} 63^{0} + \cos^{2} 63^{0}) + 1}{3 (\cos^{2} 17^{0} + \sin^{2} 17^{0}) - 2} [∵ \sin (90^{0} - θ) = \cos θ and \cos (90^{0} - θ) = \sin θ] = \frac{2 \times 1 + 1}{3 \times 1 - 2} [∵ \sin^{2} θ + \cos^{2} θ = 1] = \frac{2 + 1}{3 - 2} = \frac{3}{1} = 3$

Page No 343:

Question 10:

(sin 40° − cos 50°) = ?
(a) sin 10°
(b) cos 10°
(c) 1
(d) 0

Answer:

$(d) 0 We have: (\sin 40^{0} - \cos 50^{0}) = \sin 40^{0} - \cos (90^{0} - 40^{0}) = \sin 40^{0} - \sin 40^{0} [∵ \cos (90^{0} - θ) = \sin θ] = 0$

Page No 344:

Question 11:

(cos²37° − sin²53°) = ?
(a) $\frac{1}{3}$
(b) $\frac{2}{\sqrt{3}}$
(c) 1
(d) 0

Answer:

$(d) 0 We have: (\cos^{2} 37^{0} - \sin^{2} 53^{0}) = \cos^{2} (90^{0} - 53^{0}) - \sin^{2} 53^{0} = \sin^{2} 53^{0} - \sin^{2} 53^{0} [∵ \cos (90^{0} - θ) = \sin θ] = 0$

Page No 344:

Question 12:

(sin 43°cos 47° + cos 43°sin 47°) = ?
(a) sin 4°
(b) cos 4°
(c) 1
(d) 0

Answer:

$(c) 1 We have: (\sin 43^{0} \cos 47^{0} + \cos 43^{0} \sin 47^{0}) = \sin 43^{0} \cos (90^{0} - 43^{0}) + \cos 43^{0} \sin (90^{0} - 43^{0}) = \sin 43^{0} \sin 43^{0} + \cos 43^{0} \cos 43^{0} [∵ \cos (90^{0} - θ) = \sin θ and \sin (90^{0} - θ) = \cos θ] = \sin^{2} 43^{0} + \cos^{2} 43^{0} = 1$

Page No 344:

Question 13:

$(\frac{\sin 49 °}{\cos 41 °} - \frac{\cos 17 °}{\sin 73 °}) = ?$
(a) 1
(b) 0
(c) −1
(d) Cannot be calculated

Answer:

$(b) 0 We have: (\frac{\sin 49^{0}}{\cos 41^{0}} - \frac{\cos 17^{0}}{\sin 73^{0}}) = [\frac{\sin 49^{0}}{\cos (90^{0} - 49^{0})} - \frac{\cos (90^{0} - 73^{0})}{\sin 73^{0}}] = (\frac{\sin 49^{0}}{\sin 49^{0}} - \frac{\sin 73^{0}}{\sin 73^{0}}) [∵ \cos (90^{0} - θ) = \sin θ] = 1 - 1 = 0$

Page No 344:

Question 14:

(sin 79° cos 11° + cos 79° sin 11°) = ?
(a) $\frac{1}{\sqrt{2}}$
(b) $\frac{1}{2}$
(c) 0
(d) 1

Answer:

$(d) 1 We have: (\sin 79^{0} \cos 11^{0} + \cos 79^{0} \sin 11^{0}) = \sin 79^{0} \cos (90^{0} - 79^{0}) + \cos 79^{0} \sin (90^{0} - 79^{0}) = \sin 79^{0} \sin 79^{0} + \cos 79^{0} \cos 79^{0} [∵ \cos (90^{0} - θ) = \sin θ and \sin (90^{0} - θ) =cos θ] = \sin^{2} 79^{0} + \cos^{2} 79^{0} = 1$

Page No 344:

Question 15:

$\frac{\cot (90 ° - θ) \cdot \sin (90 ° - θ)}{sinθ} + \frac{\cot 40 °}{\tan 50 °} - (\cos^{2} 20 ° + \cos^{2} 70 °) = ?$
(a) 3
(b) 2
(c) 1
(d) 0

Answer:

$(c) 1 W e have: [\frac{\cot (90^{0} - θ) . \sin (90^{0} - θ)}{\sin θ} + \frac{\cot 40^{0}}{\tan 50^{0}} - (\cos^{2} 20^{0} + \cos^{2} 70^{0})] = [\frac{\tan θ . \cos θ}{\sin θ} + \frac{\cot (90^{0} - 50^{0})}{\tan 50^{0}} - {\cos^{2} (90^{0} - 70^{0}) + \cos^{2} 70^{0}}] [\begin{array}{l} ∵ \cot (90^{0} - θ) = \tan θ and \\ \sin (90^{0} - θ) = \cos θ \end{array}] = [\frac{\frac{\sin θ}{c o s θ} . c o s θ}{\sin θ} + \frac{\tan 50^{0}}{\tan 50^{0}} - (\sin^{2} 70^{0} + \cos^{2} 70^{0})] [∵ cos (90^{0} - θ) = \sin θ] = (\frac{\sin θ}{\sin θ} + 1 - 1) = 1 + 1 - 1 = 1$

Page No 344:

Question 16:

$\frac{2 \tan^{2} 30 ° \sec^{2} 52 ° \sin^{2} 38 °}{({cosec}^{2} 70 ° - \tan^{2} 20 °)} = ?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 2
(d) $\frac{1}{2}$

Answer:

(b) $\frac{2}{3}$

$W e have: [\frac{2 \tan^{2} 30^{0} \sec^{2} 52^{0} \sin^{2} 38^{0}}{\cos e c^{2} 70^{0} - \tan^{2} 20^{0}}] = [\frac{2 \times {(\frac{1}{\sqrt{3}})}^{2} \sec^{2} 52^{0} {\sin^{2} (90^{0} - 52^{0})}}{{\cos {ec}^{2} (90^{0} - 20^{0})} - \tan^{2} 20^{0}}] = [\frac{2}{3} \times \frac{\sec^{2} 52^{0} . \cos^{2} 52^{0}}{\sec^{2} 20^{0} - \tan^{2} 20^{0}}] [∵ \sin (90^{0} - θ) = \cos θ and \cos ec (90^{0} - θ) = \sec θ] = \frac{2}{3} \times \frac{1}{1} [∵ \sec^{2} θ - \tan^{2} θ = 1] = \frac{2}{3}$

Page No 344:

Question 17:

$\{\frac{(\sin^{2} 22 ° + \sin^{2} 68 °)}{(\cos^{2} 22 ° + \cos^{2} 68 °)} + \sin^{2} 63 ° + \cos 63 ° \sin 27\} = ?$
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

$(c) 2 We have: [\frac{\sin^{2} 22^{0} + \sin^{2} 68^{0}}{\cos^{2} 22^{0} + \cos^{2} 68} + \sin^{2} 63^{0} + \cos 63^{0} \sin 27^{0}] = [\frac{\sin^{2} 22^{0} + \sin^{2} (90^{0} - 22^{0})}{\cos^{2} (90^{0} - 68^{0}) + \cos^{2} 68^{0}} + \sin^{2} 63^{0} + \cos 63^{0} {\sin (90^{0} - 63^{0})}] = [\frac{\sin^{2} 22^{0} + \cos^{2} 22^{0}}{\sin^{2} 68^{0} + \cos^{2} 68^{0}} + \sin^{2} 63^{0} + \cos 63^{0} \cos 63^{0}] [∵ \sin (90^{0} - θ) = \cos θ and \cos (90^{0} - θ) = \sin θ] = [\frac{1}{1} + \sin^{2} 63^{0} + \cos^{2} 63^{0}] [∵ \sin^{2} θ + \cos^{2} θ = 1] = 1 + 1 = 2$

Page No 344:

Question 18:

(cosec²57° − tan²33°) = ?
(a) 0
(b) 1
(c) 2
(d) None of these

Answer:

$(b) 1 We have: (\cos {ec}^{2} 57^{0} - \tan^{2} 33^{0}) = [\cos {ec}^{2} (90^{0} - 33^{0}) - \tan^{2} 33^{0}] = (\sec^{2} 33^{0} - \tan^{2} 33^{0}) [∵ \cos ec (90^{0} - θ) = \sec θ] = 1 [∵ \sec^{2} θ - \tan^{2} θ = 1]$

Page No 344:

Question 19:

(cos²28° − sin²62°) = ?
(a) 0
(b) 1
(c) 2
(d) None of these

Answer:

$(a) 0 We have: (\cos^{2} 28^{0} - \sin^{2} 62^{0}) = [\cos^{2} (90^{0} - 62^{0}) - \sin^{2} 62^{0}] = (\sin^{2} 62^{0} - \sin^{2} 62^{0}) [∵ \cos (90^{0} - θ) = \sin θ] = 0$

Page No 344:

Question 20:

(sec²10° − cot²80°) = ?
(a) 1
(b) 0
(c) $\frac{3}{2}$
(d) None of these

Answer:

$(a) 1 We have: (\sec^{2} 10^{0} - \cot^{2} 80^{0}) = [\sec^{2} 10^{0} - \cot^{2} (90^{0} - 10^{0})] = (\sec^{2} 10^{0} - \tan^{2} 10^{0}) [∵ \cot (90^{0} - θ) = \tan θ] = 1 [\sec^{2} θ - \tan^{2} θ = 1]$

Page No 344:

Question 21:

cos (40° + θ) − sin (50° − θ) = ?
(a) 1
(b) 0
(c) sin 2θ
(d) None of these

Answer:

$(b) 0 We have: [\cos (40^{0} + θ) - \sin (50^{0} - θ)] = [\cos {90^{0} - (50^{0} - θ)} - \sin (50^{0} - θ)] = [\sin (50^{0} - θ) - \sin (50^{0} - θ)] [∵ \cos (90^{0} - θ) = \sin θ] = 0$

Page No 344:

Question 22:

sin (45° + θ) − cos (45° − θ) = ?
(a) 2 sin θ
(b) 2 cos θ
(c) 0
(d) 1

Answer:

$(c) 0 We have: [\sin (45^{0} + θ) - \cos (45^{0} - θ)] = [\sin {90^{0} - (45^{0} - θ)} - \cos (45^{0} - θ)] =[cos (45^{0} - θ) - \cos (45^{0} - θ)] [∵ \sin (90^{0} - θ) = c o s θ] = 0$

Page No 345:

Question 23:

cosec (75° + θ) −sec (15° − θ) = ?
(a) 2 sec θ
(b) 2 cosec θ
(c) 0
(d) 1

Answer:

$(c) 0 . . We have: [\cos ec (75^{0} + θ) - \sec (15^{0} - θ)] = [\cos ec {90^{0} - (15^{0} - θ)} - \sec (15^{0} - θ)] =sec (15^{0} - θ) - \sec (15^{0} - θ) [∵ \cos ec (90^{0} - θ) = \sec θ] = 0$

Page No 345:

Question 24:

If sin (θ + 34°) = cos θ and θ is acute, then θ = ?
(a) 56°
(b) 66°
(c) 28°
(d) 42°

Answer:

(d) 42°

$We have: \sin (θ + 34^{0}) = \cos θ = > \sin (θ + 34^{0}) = \sin (90^{0} - θ) [∵ \cos θ = \sin (90^{0} - θ)] = > θ + 34^{0} = 90^{0} - θ = > 2 θ = 56^{0} ∴ θ = 28^{0}$

Page No 345:

Question 25:

sin θ cos (90° − θ) + cos θ sin (90° − θ) = ?
(a) 0
(b) 1
(c) 2
(d) $\frac{3}{2}$

Answer:

(b) 1

$We have: \sin θ \cos (90^{0} - θ) + \cos θ \sin (90^{0} - θ) = \sin θ . \sin θ + \cos θ . \cos θ [∵ \cos (90^{0} - θ) = \sin θ and sin (90^{0} - θ) = \cos θ] = \sin^{2} θ + \cos^{2} θ = 1$

Page No 345:

Question 26:

$\frac{\cos 38 ° cosec 52 °}{\tan 18 ° \tan 35 ° \tan 60 ° \tan 72 ° \tan 55 °} = ?$
(a) $\sqrt{3}$
(b) $\frac{1}{3}$
(c) $\frac{1}{\sqrt{3}}$
(d) $\frac{2}{\sqrt{3}}$

Answer:

(c) $\frac{1}{\sqrt{3}}$

$We have: [\frac{\cos 38^{0} \cos ec 52^{0}}{\tan 18^{0} \tan 35^{0} \tan 60^{0} \tan 72^{0} \tan 55^{0}}] = [\frac{\cos 38^{0} \cos ec (90^{0} - 38^{0})}{\tan 18^{0} \tan 35^{0} \times \sqrt{3} \times \tan (90^{0} - 18^{0}) \tan (90^{0} - 35^{0})}] [\begin{array}{l} ∵ \cos ec (90^{0} - θ) = \sec θ and \\ \tan (90^{0} - θ) = \cot θ \end{array}] = [\frac{\cos 38^{0} \sec 38^{0}}{\tan 18^{0} \tan 35^{0} \times \sqrt{3} \times \cot 18^{0} \cot 35^{0}}] = [\frac{\frac{1}{\sec 38^{0}} \times \sec 38^{0}}{\frac{1}{\cot 18^{0}} \frac{1}{\cot 35^{0}} \times \sqrt{3} \cot 18^{0} \cot 35^{0}}] = \frac{1}{\sqrt{3}}$

Page No 345:

Question 27:

If sin 3A = cos (A − 10°), where 3A is an acute angle, then ∠A = ?
(a) 35°
(b) 25°
(c) 20°
(d) 45°

Answer:

$(b) 25° We have: [\sin 3 A=cos (A - 10^{0})] = > \cos (90^{0} - 3 A) = \cos (A - 10^{0}) [∵ \sin θ = \cos (90^{0} - θ)] = > 90^{0} - 3 A = A - 10^{0} = > - 4 A= - 100 = > A= \frac{10 0^{25}}{4_{1}} = > {A=25}^{0}$

Page No 345:

Question 28:

If tan 2A = cot (A − 21°), where 2A is an acute angle, then ∠A = ?
(a) 24°
(b) 27°
(c) 35°
(d) 37°

Answer:

(d) 37°

$We have: [\tan 2 A=cot (A - 21^{0})] = > \cot (90^{0} - 2 A) = \cot (A - 21^{0}) [∵ \tan θ = \cot (90^{0} - θ)] = > 90^{0} - 2 A = A - 21^{0} = > - 3 A= - 111 = > A= \frac{11 1^{37}}{3_{1}} = > {A=37}^{0}$

Page No 345:

Question 29:

If sec 5A = cosec (A − 30°), where 5A is an acute angle, then ∠A = ?
(a) 35°
(b) 25°
(c) 20°
(d) 27°

Answer:

$(c) 20° W e have: [\sec 5 A=cosec (A - 30^{0})] = > cosec (90^{0} - 5 A) = cosec (A - 30^{0}) [∵ \sec θ = cosec (90^{0} - θ)] = > 90^{0} - 5 A = A - 30^{0} = > - 6 A= - 120 = > A= \frac{12 0^{20}}{6_{1}} = > {A=20}^{0}$

Page No 345:

Question 30:

If A and B are acute angles such that sin A = cos B, then (A + B) = ?
(a) 45°
(b) 60°
(c) 90°
(d) 180°

Answer:

$(c) 90° We have: sinA=cosB = > cos (90^{0} - A) = cos B [∵ B and A are acute] = > 90^{0} - A = B = > {A+B=90}^{0} ∴ {A+B=90}^{0}$

Page No 345:

Question 31:

sec 70° sin 20° + cos 20° cosec 70° = ?
(a) 0
(b) 1
(c) −1
(d) 2

Answer:

$(d) 2 We have: {sec70}^{0} \sin 20^{0} + \cos 20^{0} \cos {ec70}^{0} = \frac{{sin20}^{0}}{{cos70}^{0}} + \frac{\cos 20^{0}}{{sin70}^{0}} = \frac{\sin 20^{0}}{\cos (90^{0} - 20^{0})} + \frac{{cos20}^{0}}{\sin (90^{0} - 20^{0})} = \frac{\sin 20^{0}}{\sin 20^{0}} + \frac{{cos20}^{0}}{{cos20}^{0}} [\begin{array}{l} ∵ \cos (90^{0} - θ) = \sin θ and \\ \sin (90^{0} - θ) = \cos θ \end{array}] = 1 + 1 = 2$

Page No 345:

Question 32:

cot (90° − θ) = ?
(a) cot θ
(b) −cot θ
(c) tan θ
(d) −tan θ

Answer:

$(c) tan θ cot (90^{0} - θ) = \tan θ$

Page No 345:

Question 33:

If cos 9 α = sin α and 9α < 90°, then tan 5 α = ?
(a) $\frac{1}{\sqrt{3}}$
(b) $\sqrt{3}$
(c) 1
(d) 0

Answer:

(c) 1

$We have: cos9 α =sin α = > cos9 α = cos (90^{0} - α) [∵ sin θ = cos (90^{0} - θ)] = > 9 α = 90^{0} - α . = > 10 α = 90^{0} = > α = \frac{9 0^{0}}{10} ∴ α = 9^{0} Again, tan5 α =tan {(5 \times 9)}^{0} {=>tan45}^{0} = 1$

Page No 345:

Question 34:

If cos(α + β) = 0, then sin(α − β) = ?
(a) sin α
(b) cos β
(c) sin 2α
(d) cos 2β

Answer:

(d) cos 2β

$We have: \cos (α + β) = 0 = > \cos (α + β) = \cos 90^{0} = > α + β = 90^{0} = > α = 90^{0} - β ... (i) Now, sin (α - β) = \sin [(90^{0} - β) - β] [U sing (i)] = \sin (90^{0} - 2 β) = \cos 2 β [∵ \sin (90^{0} - θ) = \cos θ]$

Page No 348:

Question 1:

$\frac{\cos^{2} 56 ° + \cos^{2} 34 °}{\sin^{2} 56 ° + \sin^{2} 34 °} + 3 \tan^{2} 56 ° \tan^{2} 34 ° = ?$
(a) $3 \frac{1}{2}$
(b) 4
(c) 6
(d) 5

Answer:

(b) 4

$\frac{\cos^{2} 56^{0} + \cos^{2} 34^{0}}{\sin^{2} 56^{0} + \sin^{2} 34^{0}} + 3 {tan}^{2} 56^{0} {tan}^{2} 34^{0} = \frac{{\cos (90^{0} - 34^{0})}^{2} + \cos^{2} 34^{0}}{{\sin (90^{0} - 34^{0})}^{2} + \sin^{2} 34^{0}} + 3 {\tan (90^{0} - 34^{0})}^{2} {tan}^{2} 34^{0} = \frac{\sin^{2} 34^{0} + \cos^{2} 34^{0}}{\cos^{2} 34^{0} + \sin^{2} 34^{0}} + 3 \cot^{2} 34^{0} {tan}^{2} 34^{0} [\begin{array}{l} ∵ \cos (90^{0} - θ) = \sin θ, \sin (90^{0} - θ) \\ = \cos θ and \tan (90^{0} - θ) = \cot θ \end{array}] = \frac{1}{1} + 3 \times 1 [∵ \cot θ = \frac{1}{\tan θ} and \sin^{2} θ + \cos^{2} θ = 1] =4$

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Question 2:

The value of $(\sin^{2} 30 ° \cos^{2} 45 ° + 4 \tan^{2} 30 ° + \frac{1}{2} \sin^{2} 90 ° + \frac{1}{8} \cot^{2} 60 °) = ?$
(a) $\frac{3}{8}$
(b) $\frac{5}{8}$
(c) 6
(d) 2

Answer:

(d) 2
$(\sin^{2} 30^{0} \cos^{2} 45^{0}) + 4 \tan^{2} 30^{0} + \frac{1}{2} \sin^{2} 90^{0} + \frac{1}{8} \cot^{2} 60^{0} = \frac{1}{2^{2}} \times \frac{1}{{(\sqrt{2})}^{2}} + 4 \times \frac{1}{{(\sqrt{3})}^{2}} + \frac{1}{2} \times 1^{2} + \frac{1}{8} \times \frac{1}{{(\sqrt{3})}^{2}} [\begin{array}{l} ∵ \sin 30^{0} = \frac{1}{2} and \cos 45^{0} = \frac{1}{\sqrt{2}} \\ {and tan30}^{0} = \frac{1}{2} and \cot 60^{0} = \frac{1}{\sqrt{3}} \end{array}] = \frac{1}{4} \times \frac{1}{2} + 4 \times \frac{1}{3} + \frac{1}{2} + \frac{1}{24} = \frac{1}{8} + \frac{4}{3} + \frac{1}{2} + \frac{1}{24} = \frac{3 + 32 + 12 + 1}{24} = \frac{48}{24} = 2$

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Question 3:

If cos A + cos² A = 1, then (sin² A + sin4 A) = ?
(a) $\frac{1}{2}$
(b) 2
(c) 1
(d) 4

Answer:

$\begin{array}{l} (c) 1 \\ cosA +^{2} A =1 \end{array} => \cos A {=sin}^{2} A . . . (i) Squaring both sides of (i), we get : \cos^{2} A = \sin^{4} A . . . (ii) A dding (i) and (ii), we get : \sin^{2} A + \sin^{4} A = \cos A + \cos^{2} A {=>sin}^{2} A + \sin^{4} A = 1 [∵ \cos A {+cos}^{2} A =1]$

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Question 4:

If sin $θ = \frac{\sqrt{3}}{2}$ , then (cosec θ + cot θ) = ?
(a) $(2 + \sqrt{3})$
(b) $2 \sqrt{3}$
(c) $\sqrt{2}$
(d) $\sqrt{3}$

Answer:

(d) $\sqrt{3}$
$Given: sin θ = \frac{\sqrt{3}}{2} and \cos ec θ = \frac{2}{\sqrt{3}} \cos {ec}^{2} θ - \cot^{2} θ = 1 => \cot^{2} θ = \cos {ec}^{2} θ - 1 => \cot^{2} θ = \frac{4}{3} - 1 [Given] => \cot θ = \frac{1}{\sqrt{3}} ∴ \cos ec θ + \cot θ = \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}} = \sqrt{3}$

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Question 5:

If cot $A = \frac{4}{5}$ , prove that $\frac{(\sin A + \cos A)}{(\sin A - \cos A)} = 9 .$

Answer:

$Given : cotA = \frac{4}{5} Writing \cot A = \frac{\cos A}{\sin A} and sqauring the equation, we get : \frac{\cos^{2} A}{\sin^{2} A} = \frac{16}{25} =>25 \cos^{2} A = 16 \sin^{2} A =>25 \cos^{2} A = 16 - 16 \cos^{2} A => \cos^{2} A = \frac{16}{41} => cosA = \frac{4}{\sqrt{41}} ∴ \sin^{2} A = 1 - \cos^{2} A =1 - \frac{16}{41} Now, \sin A = \sqrt{\frac{25}{41}} =>sin A = \frac{5}{\sqrt{41}} ∴ LHS= \frac{\sin A + cosA}{\sin A - cosA} = \frac{\frac{5}{\sqrt{41}} + \frac{4}{\sqrt{41}}}{\frac{5}{\sqrt{41}} - \frac{4}{\sqrt{41}}} = \frac{9}{1} =9 = RHS$

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Question 6:

If 2x = sec A and $\frac{2}{x}$ = tan A, prove that $(x^{2} - \frac{1}{x^{2}}) = \frac{1}{4} .$

Answer:

$Given: 2 x = secA => x = \frac{secA}{2} . . . (i) and \frac{2}{x} = tanA => \frac{1}{x} = \frac{tanA}{2} . . . (ii) ∴ x + \frac{1}{x} = \frac{secA}{2} + \frac{tanA}{2} [∵ From (i) and (ii)] Also, x - \frac{1}{x} = \frac{secA}{2} - \frac{tanA}{2} ∴ (x + \frac{1}{x}) (x - \frac{1}{x}) = (\frac{secA}{2} + \frac{tanA}{2}) (\frac{secA}{2} - \frac{tanA}{2}) => x^{2} - \frac{1}{x^{2}} = \frac{1}{4} (\sec^{2} A - \tan^{2} A) ∴ x^{2} - \frac{1}{x^{2}} = \frac{1}{4} \times 1 (∵ \sec^{2} A - \tan^{2} A = 1) = \frac{1}{4} Hence proved.$

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Question 7:

If $\sqrt{3}$ tan θ = 3 sin θ, prove that (sin² θ − cos² θ) = $\frac{1}{3} .$

Answer:

$Given: \sqrt{3} tanθ = 3 sinθ => \frac{\sqrt{3}}{cosθ} = 3 [∵ tanθ = \frac{sinθ}{cosθ}] => cosθ = \frac{\sqrt{3}}{3} => \cos^{2} θ = \frac{3}{9} ∴ \sin^{2} θ = 1 - \frac{3}{9} => \sin^{2} θ = \frac{6}{9} ∴ LHS= \sin^{2} θ - \cos^{2} θ = \frac{6}{9} - \frac{3}{9} [∴ \sin^{2} θ = \frac{6}{9}, \cos^{2} θ = \frac{3}{9}] = \frac{3}{9} = \frac{1}{3} =RHS H ence proved.$

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Question 8:

Prove that $\frac{(\sin^{2} 73 ° + \sin^{2} 17 °)}{(\cos^{2} 28 ° + \cos^{2} 62 °)} = 1 .$

Answer:

$\frac{(\sin^{2} 73 ° + \sin^{2} 17 °)}{(\cos^{2} 28 ° + \cos^{2} 62 °)} = 1 .$

$LHS= \frac{\sin^{2} 73^{0} + \sin^{2} 17^{0}}{\cos^{2} 28^{0} + \cos^{2} 62^{0}} = \frac{{[\sin (90^{0} - 17^{0})]}^{2} + \sin^{2} 17^{0}}{{[\cos (90^{0} - 62^{0})]}^{2} + \cos^{2} 62^{0}} = \frac{\cos^{2} 17^{0} + \sin^{2} 17^{0}}{\sin^{2} 62^{0} + \cos^{2} 62^{0}} = \frac{1}{1} [∵ \sin^{2} θ + \cos^{2} θ = 1] =1=RHS$

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Question 9:

If 2 sin 2θ = $\sqrt{3}$ , prove that θ = 30°.

Answer:

$2sin (2 θ) = \sqrt{3} = > sin (2 θ) = \frac{\sqrt{3}}{2} = > sin (2 θ) = \sin (60^{0}) = > 2 θ = 60^{0} = > θ = \frac{60^{0}}{2} = > θ = 30^{0}$

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Question 10:

Prove that $\sqrt{\frac{1 + \cos A}{1 - \cos A}}$ = (cosec A + cot A).

Answer:

$\sqrt{\frac{1 + \cos A}{1 - \cos A}}$ = (cosec A + cot A).

$LHS = \sqrt{\frac{1 + \cos A}{1 - \cos A}} Multiplying the numerator and denominator by (1 + \cos A), we have : \sqrt{\frac{{(1 + \cos A)}^{2}}{(1 - \cos A) (1 + \cos A)}} = \sqrt{\frac{{(1 + \cos A)}^{2}}{1 - \cos^{2} A}} = \frac{1 + \cos A}{\sqrt{\sin^{2} A}} = \frac{1 + \cos A}{\sin A} = \frac{1}{\sin A} + \frac{\cos A}{\sin A} = \cos ec A + \cot A = RHS Hence proved .$

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Question 11:

If cosec θ + cot θ = p, prove that cos θ = $\frac{(p^{2} - 1)}{(p^{2} + 1)} .$

Answer:

$\cos ec θ + \cot θ = p = > \frac{1}{\sin θ} + \frac{\cos θ}{\sin θ} = p = > \frac{1 + \cos θ}{\sin θ} = p Squaring both sides, we get: {(\frac{1 + \cos θ}{\sin θ})}^{2} = p^{2} = > \frac{{(1 + \cos θ)}^{2}}{\sin^{2} θ} = p^{2} = > \frac{{(1 + \cos θ)}^{2}}{1 - \cos^{2} θ} = p^{2} = > \frac{{(1 + \cos θ)}^{2}}{(1 + \cos θ) (1 - \cos θ)} = p^{2} = > \frac{(1 + \cos θ)}{(1 - \cos θ)} = p^{2} = > 1 + \cos θ = p^{2} (1 - \cos θ) = > 1 + \cos θ = p^{2} - p^{2} \cos θ = > \cos θ (1 + p^{2}) = p^{2} - 1 = > \cos θ = \frac{p^{2} - 1}{p^{2} + 1} Hence proved .$

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Question 12:

Prove that (cosec A − cot A)² = $\frac{(1 - \cos A)}{(1 + \cos A)} .$

Answer:

(cosec A − cot A)² = $\frac{(1 - \cos A)}{(1 + \cos A)} .$

$LHS= {(\cos ec A - \cot A)}^{2} = {(\frac{1}{\sin A} - \frac{\cos A}{\sin A})}^{2} = {(\frac{1 - \cos A}{\sin A})}^{2} = \frac{{(1 - \cos A)}^{2}}{\sin^{2} A} = \frac{{(1 - \cos A)}^{2}}{1 - \cos^{2} A} [∵ \sin^{2} θ + \cos^{2} θ = 1] = \frac{(1 - \cos A) (1 - \cos A)}{(1 - \cos A) (1 + \cos A)} = \frac{(1 - \cos A)}{(1 + \cos A)} = RHS Hence proved .$

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Question 13:

If 5 cot θ = 3, find the value of $(\frac{5 sinθ - 3 cosθ}{4 sinθ + 3 cosθ}) .$

Answer:

$Given: 5cot θ =3 => \frac{5 \cos θ}{\sin θ} = 3 [∵ cot θ = \frac{\cos θ}{\sin θ}] => 5 \cos θ = 3 \sin θ Squaring both sides, we get: 25 \cos^{2} θ = 9 \sin^{2} θ =>25 \cos^{2} θ = 9 - 9 \cos^{2} θ [∵ \sin^{2} θ + \cos^{2} θ = 1] =>34 \cos^{2} θ = 9 => \cos θ = \sqrt{\frac{9}{34}} => \cos θ = \frac{3}{\sqrt{34}} Again, \sin^{2} θ = 1 - \cos^{2} θ => \sin^{2} θ = \frac{34 - 9}{34} = \frac{25}{34} => \sin θ = \frac{5}{\sqrt{34}} ∴ L HS= (\frac{5 \sin θ - 3 \cos θ}{4 \sin θ + 3 \cos θ}) = \frac{5 \times \frac{5}{\sqrt{34}} - 3 \times \frac{3}{\sqrt{34}}}{4 \times \frac{5}{\sqrt{34}} + 3 \times \frac{3}{\sqrt{34}}} [∵ \cos θ = \frac{3}{\sqrt{34}}, \sin θ = \frac{5}{\sqrt{34}}] = \frac{25 - 9}{20 + 9} = \frac{16}{29}$

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Question 14:

Prove that (sin 32° cos 58° + cos 32° sin 58°) = 1.

Answer:

(sin 32° cos 58° + cos 32° sin 58°) = 1

${LHS=sin32}^{0} \cos 58^{0} + \cos 32^{0} {sin58}^{0} =sin (90^{0} - 58^{0}) \cos 58^{0} + \cos (90^{0} - 58^{0}) {sin58}^{0} = \cos 58^{0} \times \cos 58^{0} + {sin58}^{0} \times {sin58}^{0} [\begin{array}{l} ∵ sin (90^{0} - θ) = \cos θ, \\ \cos (90^{0} - θ) = \cos θ \end{array}] = \cos^{2} 58^{0} + {sin}^{2} 58^{0} =1 [∵ {sin}^{2} θ + \cos^{2} θ = 1] =RHS$

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Question 15:

If x = a sin θ + b cos 0 and y = a cos 0 b sin θ, prove that $x^{2} + y^{2} = a^{2} + b^{2} .$

Answer:

$Given: x = a \sin θ + b \cos θ Squaring both sides, we get: x^{2} = a^{2} \sin^{2} θ + 2 a b \sin θ \cos θ + b^{2} \cos^{2} θ . . . (i) Also, y = a \cos θ - b \sin θ Squaring both sides, we get: y^{2} = a^{2} \cos^{2} θ - 2 a b \sin θ \cos θ + b^{2} \sin^{2} θ . . . (ii) ∴ LHS= x^{2} + y^{2} = a^{2} \sin^{2} θ + 2 a b \sin θ \cos θ + b^{2} \cos^{2} θ + a^{2} \cos^{2} θ - 2 a b \sin θ \cos θ + b^{2} \sin^{2} θ [using (i) and (ii)] = a^{2} (\sin^{2} θ + \cos^{2} θ) + b^{2} (\sin^{2} θ + \cos^{2} θ) = a^{2} + b^{2} [∵ \sin^{2} θ + \cos^{2} θ = 1] =RHS Hence proved.$

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Question 16:

Prove that $\frac{(1 + sinθ)}{(1 - sinθ)}$ = (sec θ + tan θ)².

Answer:

$\frac{(1 + sinθ)}{(1 - sinθ)}$ = (sec θ + tan θ)²

$LHS= \frac{(1 + \sin θ)}{(1 - \sin θ)} Multiplying the numerator and denominator by (1 + \sin θ), we get : \frac{{(1 + \sin θ)}^{2}}{1 - \sin^{2} θ} = \frac{1 + 2 \sin θ + \sin^{2} θ}{\cos^{2} θ} [∵ \sin^{2} θ + \cos^{2} θ = 1] {=sec}^{2} θ + 2 \times \frac{\sin θ}{\cos θ} \times sec θ + \tan^{2} θ {=sec}^{2} θ + 2 \times \tan θ \times sec θ + \tan^{2} θ = {(sec θ +tan θ)}^{2} =RHS Hence proved .$

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Question 17:

Prove that $\frac{1}{(secθ - tanθ)} - \frac{1}{cosθ} = \frac{1}{cosθ} - \frac{1}{(secθ + tanθ)}$ .

Answer:

$\frac{1}{(secθ - tanθ)} - \frac{1}{cosθ} = \frac{1}{cosθ} - \frac{1}{(secθ + tanθ)}$

$LHS= \frac{1}{secθ - tanθ} - \frac{1}{cosθ} = \frac{(secθ + tanθ)}{(secθ - tanθ) (secθ + tanθ)} - secθ (\begin{array}{l} Multipying the numerator and \\ denominator by (secθ + tanθ) \end{array}) = \frac{secθ + tanθ}{\sec^{2} θ - \tan^{2} θ} - secθ =sec θ +tan θ - secθ [∵ \sec^{2} θ - \tan^{2} θ = 1] = tanθ RHS= \frac{1}{cosθ} - \frac{1}{secθ + tanθ} =sec θ - \frac{(secθ - tanθ)}{\sec^{2} θ - \tan^{2} θ} (\begin{array}{l} Multipying the numerator and \\ denomenator by (secθ - tanθ) \end{array}) = secθ + tanθ - secθ [∵ \sec^{2} θ - \tan^{2} θ = 1] = tanθ ∴ LHS=RHS Hence Proved$

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Question 18:

Prove that $\frac{(\sin A - 2 \sin^{3} A)}{(2 \cos^{3} A - \cos A)} = \tan A .$

Answer:

$LHS= \frac{(\sin A - 2 \sin^{3} A)}{(2 \cos^{3} A - \cos A)} = \frac{\sin A (1 - 2 \sin^{2} A)}{\cos A (2 \cos^{2} A - 1)} = \tan A {\frac{(\sin^{2} A + \cos^{2} A - 2 \sin^{2} A)}{2 \cos^{2} A - \sin^{2} A - \cos^{2} A}} [∵ \sin^{2} A + \cos^{2} A = 1] = \tan A {\frac{(\cos^{2} A - \sin^{2} A)}{(\cos^{2} A - \sin^{2} A)}} = \tan A =RHS$

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Question 19:

Prove that $\frac{\tan A}{(1 - \cot A)} + \frac{\cot A}{(1 - \tan A)} = (1 + \tan A + \cot A) .$

Answer:

$LHS= \frac{\tan A}{(1 - \cot A)} + \frac{\cot A}{(1 - \tan A)} = \frac{\tan A}{(1 - \cot A)} + \frac{\cot^{2} A}{(\cot A - 1)} [∵ \tan A = \frac{1}{\cot A}] = \frac{\tan A}{(1 - \cot A)} - \frac{\cot^{2} A}{(1 - \cot A)} = \frac{\tan A - \cot^{2} A}{(1 - \cot A)} = \frac{(\frac{1}{\cot A}) - \cot^{2} A}{(1 - \cot A)} = \frac{1 - \cot^{3} A}{\cot A (1 - \cot A)} = \frac{(1 - \cot A) (1 + \cot A + \cot^{2} A)}{\cot A (1 - \cot A)} [∵ a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})] = \frac{1}{\cot A} + \frac{\cot^{2} A}{\cot A} + \frac{\cot A}{\cot A} =1+ \tan A + \cot A =RHS Hence proved$

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Question 20:

If sec 5 A = cosec (A − 36) and 5 A is an acute angle, find the value of A.

Answer:

$Given: sec5 A = \cos e c (A - 36^{0}) => cosec (90^{0} - 5 A) = \cos e c (A - 36^{0}) [∵ \cos e c (90^{0} - θ) = \sec θ] => 90^{0} - 5 A = A - 36^{0} =>6 A = 90^{0} + 36^{0} => 6 A = 126^{0} => A = 21^{0}$

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