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

Share with your friends

RS Aggarwal 2015 Solutions for Class 10 Math Chapter 8 Trigonometric Ratios Of Complementary Angles are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios Of Complementary Angles are extremely popular among class 10 students for Math Trigonometric Ratios Of Complementary Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2015 Book of class 10 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2015 Solutions. All RS Aggarwal 2015 Solutions for class 10 Math are prepared by experts and are 100% accurate.

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:

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

Page No 341:

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:

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

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

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

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

Page No 341:

Question 9:

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

Answer:

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

Page No 341:

Question 10:

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

Answer:

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

Page No 341:

Question 11:

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

Answer:

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

Page No 341:

Question 12:

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

Answer:

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

Page No 341:

Question 13:

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

Answer:

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

Page No 341:

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:

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

Page No 341:

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

Page No 341:

Question 16:

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

Answer:

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

Page No 341:

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:

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

Page No 341:

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:

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

Page No 342:

Question 19:

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

Answer:

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

Page No 342:

Question 20:

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

Answer:

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

Page No 342:

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:

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

Page No 342:

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:

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

Page No 342:

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:

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

Page No 342:

Question 24:

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

Answer:

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

Page No 342:

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:

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

Page No 342:

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 cos θ
(b) 2 sin 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) cos β
(b) cos 2β
(c) sin α
(d) sin 2α

Answer:

(b) 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

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

Page No 348:

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

Page No 348:

Question 3:

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

Answer:

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

Page No 348:

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

Page No 348:

Question 5:

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

Answer:

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

Page No 348:

Question 6:

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

Answer:

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

Page No 348:

Question 7:

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

Answer:

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

Page No 349:

Question 8:

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

Answer:

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

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

Page No 349:

Question 9:

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

Answer:

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

Page No 349:

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).

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

Page No 349:

Question 11:

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

Answer:

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

Page No 349:

Question 12:

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

Answer:

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

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

Page No 349:

Question 13:

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

Answer:

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

Page No 349:

Question 14:

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

Answer:

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

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

Page No 349:

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:

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

Page No 349:

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

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

Page No 349:

Question 17:

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

Answer:

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

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

Page No 349:

Question 18:

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

Answer:

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

Page No 349:

Question 19:

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

Answer:

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

Page No 349:

Question 20:

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

Answer:

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

View NCERT Solutions for all chapters of Class 10

Login or Create a free account