RS Aggarwal 2018 Solutions for Class 10 Math Chapter 7 - Trigonometric Ratios Of Complementary Angles

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Page No 312:

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 312:

Question 2:

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 313:

Question 3:

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 313:

Question 4:

Prove that:

$(i) \frac{\sin 70 °}{\cos 20 °} + \frac{cosec 20 °}{\sec 70 °} - 2 \cos 70 ° cosec 20 ° = 0 (ii) \frac{\cos 80 °}{\sin 10 °} + \cos 59 ° cosec 31 ° = 2 (iii) \frac{2 \sin 68 °}{\cos 22 °} - \frac{2 \cot 15 °}{5 \tan 75 °} - \frac{3 \tan 45 ° \tan 20 ° \tan 40 ° \tan 50 ° \tan 70 °}{5} = 1 (iv) \frac{\sin 18 °}{\cos 72 °} + \sqrt{3} (\tan 10 ° \tan 30 ° \tan 40 ° \tan 50 ° \tan 80 °) = 2 [CBSE 2008] (v) \frac{7 \cos 55 °}{3 \sin 35 °} - \frac{4 (\cos 70 ° cosec 20 °)}{3 (\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °)} = 1$

Answer:

$(i) LHS = \frac{\sin 70 °}{\cos 20 °} + \frac{cosec 20 °}{\sec 70 °} - 2 \cos 70 ° cosec 20 ° = \frac{\sin 70 °}{\sin (90 ° - 20 °)} + \frac{\sec (90 ° - 20 °)}{\sec 70 °} - 2 \cos 70 ° \sec (90 ° - 20 °) = \frac{\sin 70 °}{\sin 70 °} + \frac{\sec 70 °}{\sec 70 °} - 2 \cos 70 ° \sec 70 ° = 1 + 1 - 2 \times \cos 70 ° \times \frac{1}{\cos 70 °} = 2 - 2 = 0 = RHS$

$(ii) LHS = \frac{\cos 80 °}{\sin 10 °} + \cos 59 ° cosec 31 ° = \frac{\cos 80 °}{\cos (90 ° - 10 °)} + \sin (90 ° - 59 °) cosec 31 ° = \frac{\cos 80 °}{\cos 80 °} + \sin 31 ° cosec 31 ° = 1 + \sin 31 ° \times \frac{1}{\sin 31 °} = 1 + 1 = 2 = RHS$

$(iii) LHS = \frac{2 \sin 68 °}{\cos 22 °} - \frac{2 \cot 15 °}{5 \tan 75 °} - \frac{3 \tan 45 ° \tan 20 ° \tan 40 ° \tan 50 ° \tan 70 °}{5} = \frac{2 \sin 68 °}{\sin (90 ° - 22 °)} - \frac{2 \cot 15 °}{5 \cot (90 ° - 75 °)} - \frac{3 \times 1 \times \cot (90 ° - 20 °) \times \cot (90 ° - 40 °) \times \tan 50 ° \times \tan 70 °}{5} = \frac{2 \sin 68 °}{\sin 68 °} - \frac{2 \cot 15 °}{5 \cot 15 °} - \frac{3 \cot 70 ° \cot 50 ° \tan 50 ° \tan 70 °}{5} = 2 - \frac{2}{5} - \frac{3 \times \frac{1}{\tan 70 °} \times \frac{1}{\tan 50 °} \times \tan 50 ° \times \tan 70 °}{5} = 2 - \frac{2}{5} - \frac{3}{5} = \frac{10 - 2 - 3}{5} = \frac{5}{5} = 1 = RHS$

$(iv) LHS = \frac{\sin 18 °}{\cos 72 °} + \sqrt{3} (\tan 10 ° \tan 30 ° \tan 40 ° \tan 50 ° \tan 80 °) = \frac{\sin 18 °}{\sin (90 ° - 72 °)} + \sqrt{3} [\cot (90 ° - 10 °) \times \frac{1}{\sqrt{3}} \times \cot (90 ° - 40 °) \times \tan 50 ° \times \tan 80 °] = \frac{\sin 18 °}{\sin 18 °} + \sqrt{3} (\frac{\cot 80 ° \times \cot 50 ° \times \tan 50 ° \times \tan 80 °}{\sqrt{3}}) = 1 + (\frac{1}{\tan 80 °} \times \frac{1}{\tan 50 °} \times \tan 50 ° \times \tan 80 °) = 1 + 1 = 2 = RHS$

$(v) LHS = \frac{7 \cos 55 °}{3 \sin 35 °} - \frac{4 (\cos 70 ° cosec 20 °)}{3 (\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °)} = \frac{7 \cos 55 °}{3 \cos (90 ° - 35 °)} - \frac{4 [\sin (90 ° - 70 °) cosec 20 °]}{3 [\cot (90 ° - 5 °) \times \cot (90 ° - 25 °) \times 1 \times \tan 65 ° \times \tan 85 °]} = \frac{7 \cos 55 °}{3 \cos 55 °} - \frac{4 (\sin 20 ° cosec 20 °)}{3 (\cot 85 ° \cot 65 ° \tan 65 ° \tan 85 °)} = \frac{7}{3} - \frac{4 (\sin 20 ° \times \frac{1}{\sin 20 °})}{3 (\frac{1}{\tan 85 °} \times \frac{1}{\tan 65 °} \times \tan 65 ° \times \tan 85 °)} = \frac{7}{3} - \frac{4}{3} = \frac{3}{3} = 1 = RHS$

Page No 313:

Question 5:

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 313:

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 314:

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

Answer:

$In ∆ ABC, A + B + C = 180 ° \Rightarrow A + C = 180 ° - B . . . . . (i) Now, LHS = \tan (\frac{C + A}{2}) = \tan (\frac{180 ° - B}{2}) [Using (i)] = \tan (90 ° - \frac{B}{2}) = \cot \frac{B}{2} = RHS$

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Question 10:

$If \cos 2 θ = \sin 4 θ, where 2 θ and 4 θ are acute angles, then find the value of θ .$

Answer:

$We have, \cos 2 θ = \sin 4 θ \Rightarrow \sin (90 ° - 2 θ) = \sin 4 θ Comparing both sides, we get 90 ° - 2 θ = 4 θ \Rightarrow 2 θ + 4 θ = 90 ° \Rightarrow 6 θ = 90 ° \Rightarrow θ = \frac{90 °}{6} ∴ θ = 15 °$
Hence, the value of $θ$ is $15 ° .$

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Question 11:

If sec2A = cosec(A $-$ 42 $°$ ), where 2A is an acute angle, then find the value of A. [CBSE2008]

Answer:

We have,

$\sec 2 A = cosec (A - 42 °) \Rightarrow cosec (90 ° - 2 A) = cosec (A - 42 °) Comparing both sides, we get 90 ° - 2 A = A - 42 ° \Rightarrow 2 A + A = 90 ° + 42 ° \Rightarrow 3 A = 132 ° \Rightarrow A = \frac{132 °}{3} ∴ A = 44 °$
Hence, the value of A is $44 °$ .

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Question 12:

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 13:

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 14:

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 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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