Rs Aggarwal 2019 for Class 10 Math Chapter 12 - Trigonometric Ratios Of Some Complemantary Angles

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Rs Aggarwal 2019 Solutions for Class 10 Math Chapter 12 Trigonometric Ratios Of Some Complemantary Angles are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios Of Some Complemantary Angles are extremely popular among Class 10 students for Math Trigonometric Ratios Of Some Complemantary Angles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 10 Math Chapter 12 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 563:

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 563:

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 564:

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 564:

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 564:

Answer:

$\begin{array}{l} (i) LH S = \sin θ \cos (90^{0} - θ) + \sin (90^{0} - θ) \cos θ \end{array} \begin{array}{l} = sinθsinθ + cosθcosθ \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} Hence proved . \begin{array}{l} (ii) LHS = \frac{sinθ}{\cos (90^{0} - θ)} + \frac{cosθ}{\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 = RHS \end{array} Hence proved \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 564:

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 565:

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 565:

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 565:

Answer:

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

Page No 565:

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 $15 ° .$

Page No 565:

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 $44 °$ .

Page No 565:

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 565:

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 565:

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 565:

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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