Rs Aggarwal 2020 2021 for Class 10 Maths Chapter 12 - Trigonometric Ratios Of Some Complemantary Angles

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Page No 590:

Answer:

(i) $\frac{\sin 26 °}{\cos 64 °}$

$\frac{\sin 26 °}{\cos 64 °} = \frac{\sin (90 ° - 64 °)}{\cos 64 °} = \frac{\cos 64 °}{\cos 64 °} (∵ \sin (90 ° - θ) = \cos θ) = 1 Hence, \frac{\sin 26 °}{\cos 64 °} = 1$

$\begin{array}{l} (ii) \frac{\sec 11^{\circ}}{cosec 79^{\circ}} \end{array} \begin{array}{l} = \frac{\sec (90^{\circ} - 79^{\circ})}{cosec 79^{\circ}} \end{array} \begin{array}{l} = \frac{cosec 79^{\circ}}{cosec 79^{\circ}} \end{array} [∵ s e c (90 - θ) = \cos e c θ] \begin{array}{l} = 1 \end{array}$

(iii) $\frac{\tan 65 °}{\cot 25 °}$

$\frac{\tan 65 °}{\cot 25 °} = \frac{\tan (90 ° - 25 °)}{\cot 25 °} = \frac{\cot 25 °}{\cot 25 °} (∵ \tan (90 ° - θ) = \cot θ) = 1 Hence, \frac{\tan 65 °}{\cot 25 °} = 1$

(iv) $\frac{\cos 37 °}{\sin 53 °}$

$\frac{\cos 37 °}{\sin 53 °} = \frac{\cos (90 ° - 53 °)}{\sin 53 °} = \frac{\sin 53 °}{\sin 53 °} (∵ \cos (90 ° - θ) = \sin θ) = 1 Hence, \frac{\cos 37 °}{\sin 53 °} = 1$

$\begin{array}{l} (v) \frac{cosec 42^{\circ}}{\sec 48^{\circ}} \end{array} \begin{array}{l} = \frac{cosec (90^{\circ} - 48^{\circ})}{\sec 48^{\circ}} \end{array} \begin{array}{l} = \frac{\sec 48^{\circ}}{\sec 48^{\circ}} [∵ s e c (90 - θ) = \cos e c θ] \end{array} \begin{array}{l} = 1 \end{array}$

(vi) $\frac{\cot 34 °}{\tan 56 °}$

$\frac{\cot 34 °}{\tan 56 °} = \frac{\cot (90 ° - 56 °)}{\tan 56 °} = \frac{\tan 56 °}{\tan 56 °} (∵ \cot (90 ° - θ) = \tan θ) = 1 Hence, \frac{\cot 34 °}{\tan 56 °} = 1$

Page No 590:

Answer:

$\begin{array}{l} (i) {LHS=cos81}^{0} - \sin 9^{0} \end{array} \begin{array}{l} =cos (90^{0} - 9^{0}) - \sin 9^{0} \end{array} \begin{array}{l} {=sin9}^{0} - \sin 9^{0} \end{array} \begin{array}{l} =0 \end{array} \begin{array}{l} =RHS \end{array} \begin{array}{l} (ii) {LHS=tan71}^{0} - \cot 19^{0} \end{array} \begin{array}{l} =tan (90^{0} - 19^{0}) - \cot 19^{0} \end{array} \begin{array}{l} {=cot19}^{0} - \cot 19^{0} \end{array} \begin{array}{l} =0 \end{array}$

(iii) To Prove: cosec60° − sec30° = 0

$cosec 60 ° - \sec 30 ° = cosec (90 ° - 30 °) - \sec 30 ° = \sec 30 ° - \sec 30 ° (∵ cosec (90 ° - θ) = \sec θ) = 0 Hence, cosec 60 ° - \sec 30 ° = 0 .$

(iv) To Prove: cot34° − tan56° = 0

$\cot 34 ° - \tan 56 ° = \cot (90 ° - 56 °) - \tan 56 ° = \tan 56 ° - \tan 56 ° (∵ \cot (90 ° - θ) = \tan θ) = 0 Hence, \cot 34 ° - \tan 56 ° = 0 .$

$\begin{array}{l} (v) {LHS=sin}^{2} 48^{0} + \sin^{2} 42^{0} \end{array} \begin{array}{l} {=sin}^{2} (90^{0} - 42^{0}) + \sin^{2} 42^{0} \end{array} \begin{array}{l} {=cos}^{2} 42^{0} + \sin^{2} 42^{0} \end{array} \begin{array}{l} =1 \end{array} \begin{array}{l} =RHS \end{array}$

(vi) To Prove: cos²72° + cos²18° = 1

$\cos^{2} 72 ° + \cos^{2} 18 ° = {(\cos (90 ° - 18 °))}^{2} + \cos^{2} 18 ° = \sin^{2} 18 ° + \cos^{2} 18 ° (∵ \cos (90 ° - θ) = \sin θ) = 1 (using the identity : \cos^{2} θ + \sin^{2} θ = 1) Hence, \cos^{2} 72 ° + \cos^{2} 18 ° = 1 .$

Page No 590:

Answer:

(i) To prove: sin²12° + sin² 78° = 1

$\sin^{2} 12 ° + \sin^{2} 78 ° = {(\sin (90 ° - 78 °))}^{2} + \sin^{2} 78 ° = \cos^{2} 78 ° + \sin^{2} 78 ° (∵ \sin (90 ° - θ) = \cos θ) = 1 (using the identity : \cos^{2} θ + \sin^{2} θ = 1) Hence, \sin^{2} 12 ° + \sin^{2} 78 ° = 1 .$

(ii) To prove: sec²29° – cot²61° = 1

$\sec^{2} 29 ° - \cot^{2} 61 ° = \sec^{2} 29 ° - {(\cot (90 ° - 29 °))}^{2} = \sec^{2} 29 ° - \tan^{2} 29 ° (∵ \cot (90 ° - θ) = \tan θ) = 1 (using the identity : \sec^{2} θ - \tan^{2} θ = 1) Hence, \sec^{2} 29 ° - \cot^{2} 61 ° = 1 .$

(iii) To prove: tan²56° – cot²34° = 0

$\tan^{2} 56 ° - \cot^{2} 34 ° = {(\tan (90 ° - 34 °))}^{2} - \cot^{2} 34 ° = \cot^{2} 34 ° - \cot^{2} 34 ° (∵ \tan (90 ° - θ) = \cot θ) = 0 Hence, \tan^{2} 56 ° - \cot^{2} 34 ° = 0 .$

(iv) To prove: cos²57° – sin²33° = 0

$\cos^{2} 57 ° - \sin^{2} 33 ° = {(\cos (90 ° - 33 °))}^{2} - \sin^{2} 33 ° = \sin^{2} 33 ° - \sin^{2} 33 ° (∵ \cos (90 ° - θ) = \sin θ) = 0 Hence, \cos^{2} 57 ° - \sin^{2} 33 ° = 0 .$

(v) To prove: sec²50° – cot²40° = 1

$\sec^{2} 50 ° - \cot^{2} 40 ° = \sec^{2} 50 ° - {(\cot (90 ° - 50 °))}^{2} = \sec^{2} 50 ° - \tan^{2} 50 ° (∵ \cot (90 ° - θ) = \tan θ) = 1 (using the identity : \sec^{2} θ - \tan^{2} θ = 1) Hence, \sec^{2} 50 ° - \cot^{2} 40 ° = 1 .$

(vi) To prove: cosec²72° – tan²18° = 1

${cosec}^{2} 72 ° - \tan^{2} 18 ° = {(cosec (90 ° - 18 °))}^{2} - \tan^{2} 18 ° = \sec^{2} 18 ° - \tan^{2} 18 ° (∵ cosec (90 ° - θ) = \sec θ) = 1 (using the identity : \sec^{2} θ - \tan^{2} θ = 1) Hence, {cosec}^{2} 72 ° - \tan^{2} 18 ° = 1 .$

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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) To prove: tan15° tan60° tan75° = $\sqrt{3}$

$\tan 15 ° \tan 60 ° \tan 75 ° = \tan 15 ° \tan 60 ° \tan (90 ° - 15 °) = \tan 15 ° \tan 60 ° \cot 15 ° (∵ \tan (90 ° - θ) = \cot θ) = \tan 15 ° \tan 60 ° \frac{1}{\tan 15 °} (∵ \cot θ = \frac{1}{\tan θ}) = \tan 60 ° = \sqrt{3} (∵ \tan 60 ° = \sqrt{3}) Hence, \tan 15 ° \tan 60 ° \tan 75 ° = \sqrt{3} .$
$(v) LHS = \tan 48 ° \tan 23 ° \tan 42 ° \tan 67 ° = \cot (90 ° - 48 °) \cot (90 ° - 23 °) \tan 42 ° \tan 67 ° = \cot 42 ° \cot 67 ° \tan 42 ° \tan 67 ° = \frac{1}{\tan 42 °} \times \frac{1}{\tan 67 °} \times \tan 42 ° \times \tan 67 ° = 1 = RHS$

$(vi) LHS = (\sin 72 ° + \cos 18 °) (\sin 72 ° - \cos 18 °) = (\sin 72 ° + \cos 18 °) [\cos (90 ° - 72 °) - \cos 18 °] = (\sin 72 ° + \cos 18 °) (\cos 18 ° - \cos 18 °) = (\sin 72 ° + \cos 18 °) (0) = 0 = RHS$
(vii) To prove: cosec39° cos51° + tan21° cot69° – sec²21° = 0

$cosec 39 ° \cos 51 ° + \tan 21 ° \cot 69 ° - \sec^{2} 21 ° = cosec (90 ° - 51 °) \cos 51 ° + \tan 21 ° \cot (90 ° - 21 °) - \sec^{2} 21 ° = \sec 51 ° \cos 51 ° + \tan 21 ° \tan 21 ° - \sec^{2} 21 ° (∵ \cot (90 ° - θ) = \tan θ and cosec (90 ° - θ) = \sec θ) = \sec 51 ° \cos 51 ° + \tan^{2} 21 ° - \sec^{2} 21 ° = \sec 51 ° \cos 51 ° - (\sec^{2} 21 ° - \tan^{2} 21 °) (using the identity : \sec^{2} θ - \tan^{2} θ = 1) = \sec 51 ° \cos 51 ° - 1 = \frac{1}{\cos 51 °} \cos 51 ° - 1 (∵ \sec θ = \frac{1}{\cos θ}) = 1 - 1 = 0 Hence, cosec 39 ° \cos 51 ° + \tan 21 ° \cot 69 ° - \sec^{2} 21 ° = 0 .$

Page No 590:

Answer:

(i) sin72° + cosec72°

$\sin 72 ° + cosec 72 ° = \sin (90 ° - 18 °) + cosec (90 ° - 18 °) = \cos 18 ° + \sec 18 ° (∵ \sin (90 ° - θ) = \cos θ and cosec (90 ° - θ) = \sec θ) Hence, \sin 72 ° + cosec 72 ° = \cos 18 ° + \sec 18 ° .$

(ii) cosec66° + tan66°

$cosec 66 ° + \tan 66 ° = cosec (90 ° - 24 °) + \tan (90 ° - 24 °) = \sec 24 ° + \cot 24 ° (∵ \tan (90 ° - θ) = \cot θ and cosec (90 ° - θ) = \sec θ) Hence, cosec 66 ° + \tan 66 ° = \sec 24 ° + \cot 24 ° .$

(iii) tan68° + sec68°

$\tan 68 ° + \sec 68 ° = \tan (90 ° - 22 °) + \sec (90 ° - 22 °) = \cot 22 ° + cosec 22 ° (∵ \tan (90 ° - θ) = \cot θ and \sec (90 ° - θ) = cosec θ) Hence, \tan 68 ° + \sec 68 ° = \cot 22 ° + cosec 22 ° .$

(iv) cot59° + cosec59°

$\cot 59 ° + cosec 59 ° = \cot (90 ° - 31 °) + cosec (90 ° - 31 °) = \tan 31 ° + \sec 31 ° (∵ \cot (90 ° - θ) = \tan θ and cosec (90 ° - θ) = \sec θ) Hence, \cot 59 ° + cosec 59 ° = \tan 31 ° + \sec 31 ° .$

(v) cos51° + cot49° – sec47°

$\cos 51 ° + \cot 49 ° - \sec 47 ° = \cos (90 ° - 39 °) + \cot (90 ° - 41 °) - \sec (90 ° - 43 °) = \sin 39 ° + \tan 41 ° - cosec 43 ° (∵ \cos (90 ° - θ) = \sin θ, \cot (90 ° - θ) = \tan θ and \sec (90 ° - θ) = cosec θ) Hence, \cos 51 ° + \cot 49 ° - \sec 47 ° = \sin 39 ° + \tan 41 ° - cosec 43 ° .$

(vi) sin67° + cos75°

$\sin 67 ° + \cos 75 ° = \sin (90 ° - 23 °) + \cos (90 ° - 15 °) = \cos 23 ° + \sin 15 ° (∵ \sin (90 ° - θ) = \cos θ and \cos (90 ° - θ) = \sin θ) Hence, \sin 67 ° + \cos 75 ° = \cos 23 ° + \sin 15 ° .$

Page No 591:

Answer:

Given: sin3A = cos(A – 10°)

$\sin 3 A = \cos (A - 10 °) \Rightarrow \cos (90 ° - 3 A) = \cos (A - 10 °) (∵ \sin θ = \cos (90 ° - θ)) \Rightarrow 90 ° - 3 A = A - 10 ° \Rightarrow 90 ° + 10 ° = A + 3 A \Rightarrow 4 A = 100 ° \Rightarrow A = \frac{100 °}{4} \Rightarrow A = 25 ° Hence, ∠ A = 25 ° .$

Page No 591:

Answer:

Given: tanA = cot(A + 10°)

$\tan A = \cot (A + 10 °) \Rightarrow \cot (90 ° - A) = \cot (A + 10 °) (∵ \tan θ = \cot (90 ° - θ)) \Rightarrow 90 ° - A = A + 10 ° \Rightarrow 90 ° - 10 ° = A + A \Rightarrow 2 A = 80 ° \Rightarrow A = \frac{80 °}{2} \Rightarrow A = 40 ° Hence, ∠ A = 40 ° .$

Page No 591:

Answer:

Given: cos2A = sin(A – 15°)

$\cos 2 A = \sin (A - 15 °) \Rightarrow \sin (90 ° - 2 A) = \sin (A - 15 °) (∵ \cos θ = \sin (90 ° - θ)) \Rightarrow 90 ° - 2 A = A - 15 ° \Rightarrow 90 ° + 15 ° = A + 2 A \Rightarrow 3 A = 105 ° \Rightarrow A = \frac{105 °}{3} \Rightarrow A = 35 ° Hence, ∠ A = 35 ° .$

Page No 591:

Answer:

Given: sec4A = cosec(A – 15°)

$\sec 4 A = cosec (A - 15 °) \Rightarrow cosec (90 ° - 4 A) = cosec (A - 15 °) (∵ \sec θ = cosec (90 ° - θ)) \Rightarrow 90 ° - 4 A = A - 15 ° \Rightarrow 90 ° + 15 ° = A + 4 A \Rightarrow 5 A = 105 ° \Rightarrow A = \frac{105 °}{5} \Rightarrow A = 21 ° Hence, ∠ A = 21 ° .$

Page No 591:

Answer:

Given: tan2θ = cot(θ + 60°)

$\tan 2 θ = \cot (θ + 60 °) \Rightarrow \cot (90 ° - 2 θ) = \cot (θ + 60 °) (∵ \tan θ = \cot (90 ° - θ)) \Rightarrow 90 ° - 2 θ = θ + 60 ° \Rightarrow 90 ° - 60 ° = θ + 2 θ \Rightarrow 3 θ = 30 ° \Rightarrow θ = \frac{30 °}{3} \Rightarrow θ = 10 ° Hence, the value of θ is 10 ° .$

Page No 591:

Answer:

Given: sin(θ + 36°) = cosθ

$\sin (θ + 36 °) = \cos θ \Rightarrow \cos (90 ° - (θ + 36 °)) = \cos θ (∵ \sin θ = \cos (90 ° - θ)) \Rightarrow \cos (90 ° - θ - 36 °) = \cos θ \Rightarrow 54 ° - θ = θ \Rightarrow θ + θ = 54 ° \Rightarrow 2 θ = 54 ° \Rightarrow θ = \frac{54 °}{2} \Rightarrow θ = 27 ° Hence, θ = 27 ° .$

Page No 591:

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$3 \cot 31 ° \tan 15 ° \cot 27 ° \tan 75 ° \cot 63 ° \cot 59 ° = 3 \cot (90 ° - 59 °) \tan 15 ° \cot (90 ° - 63 °) \tan 75 ° \cot 63 ° \cot 59 ° = 3 \tan 59 ° \tan 15 ° \tan 63 ° \tan 75 ° \cot 63 ° \cot 59 ° (∵ \cot (90 ° - θ) = \tan θ) = 3 \tan 59 ° \tan (90 ° - 75 °) \tan 63 ° \tan 75 ° \cot 63 ° \cot 59 ° = 3 \tan 59 ° \cot 75 ° \tan 63 ° \tan 75 ° \cot 63 ° \cot 59 ° (∵ \tan (90 ° - θ) = \cot θ) = 3 (\tan 59 ° \cot 59 °) (\tan 63 ° \cot 63 °) (\cot 75 ° \tan 75 °) = 3 (\tan 59 ° \frac{1}{\tan 59 °}) (\tan 63 ° \frac{1}{\tan 63 °}) (\frac{1}{\tan 75 °} \tan 75 °) (∵ \cot θ = \frac{1}{\tan θ}) = 3 Hence, 3 \cot 31 ° \tan 15 ° \cot 27 ° \tan 75 ° \cot 63 ° \cot 59 ° = 3 .$

Page No 591:

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$2 (\frac{\cos 58 °}{\sin 32 °}) - \sqrt{3} (\frac{\cos 38 ° cosec 52 °}{\tan 15 ° \tan 60 ° \tan 75 °}) = 2 (\frac{\cos (90 ° - 32 °)}{\sin 32 °}) - \sqrt{3} (\frac{\cos (90 ° - 52 °) cosec 52 °}{\tan 15 ° \tan 60 ° \tan 75 °}) = 2 (\frac{\sin 32 °}{\sin 32 °}) - \sqrt{3} (\frac{\sin 52 ° cosec 52 °}{\tan 15 ° \tan 60 ° \tan 75 °}) (∵ \cos (90 ° - θ) = \sin θ) = 2 (1) - \sqrt{3} (\frac{\sin 52 ° \frac{1}{\sin 52 °}}{\tan 15 ° \tan 60 ° \tan 75 °}) (∵ cosec θ = \frac{1}{\sin θ}) = 2 - \sqrt{3} (\frac{1}{\tan 15 ° \tan 60 ° \tan 75 °}) = 2 - \sqrt{3} (\frac{1}{\tan 15 ° \tan 60 ° \tan (90 ° - 15 °)}) = 2 - \sqrt{3} (\frac{1}{\tan 15 ° \tan 60 ° \cot 15 °}) (∵ \tan (90 ° - θ) = \cot θ) = 2 - \sqrt{3} (\frac{1}{\tan 15 ° \tan 60 ° \frac{1}{\tan 15 °}}) (∵ \cot θ = \frac{1}{\tan θ}) = 2 - \sqrt{3} (\frac{1}{\tan 60 °}) = 2 - \sqrt{3} (\frac{1}{\sqrt{3}}) (∵ \tan 60 ° = \sqrt{3}) = 2 - 1 = 1 Hence, 2 (\frac{\cos 58 °}{\sin 32 °}) - \sqrt{3} (\frac{\cos 38 ° cosec 52 °}{\tan 15 ° \tan 60 ° \tan 75 °}) = 1 .$

Page No 591:

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$\{\frac{\cos 70 °}{\sin 20 °} + \frac{\cos 55 ° cosec 35 °}{\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °}\} = \{\frac{\cos (90 ° - 20 °)}{\sin 20 °} + \frac{\cos (90 ° - 35 °) cosec 35 °}{\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °}\} = \{\frac{\sin 20 °}{\sin 20 °} + \frac{\sin 35 ° cosec 35 °}{\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °}\} (∵ \cos (90 ° - θ) = \sin θ) = \{1 + \frac{\sin 35 ° \frac{1}{\sin 35 °}}{\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °}\} (∵ cosec θ = \frac{1}{\sin θ}) = \{1 + \frac{1}{\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °}\} = \{1 + \frac{1}{\tan 5 ° \tan 25 ° \tan 45 ° \tan (90 ° - 25 °) \tan (90 ° - 5 °)}\} = \{1 + \frac{1}{\tan 5 ° \tan 25 ° \tan 45 ° \cot 25 ° \cot 5 °}\} (∵ \tan (90 ° - θ) = \cot θ) = \{1 + \frac{1}{\tan 5 ° \tan 25 ° \tan 45 ° \frac{1}{\tan 25 °} \frac{1}{\tan 5 °}}\} (∵ \cot θ = \frac{1}{\tan θ}) = \{1 + \frac{1}{\tan 45 °}\} = \{1 + \frac{1}{1}\} (∵ \tan 45 ° = 1) = 1 + 1 = 2 Hence, \{\frac{\cos 70 °}{\sin 20 °} + \frac{\cos 55 ° cosec 35 °}{\tan 5 ° \tan 25 ° \tan 45 ° \tan 65 ° \tan 85 °}\} = 2 .$

Page No 591:

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$\{\frac{\sin^{2} 22 ° + \sin^{2} 68 °}{\cos^{2} 22 ° + \cos^{2} 68 °} + \sin^{2} 63 ° + \cos 63 ° \sin 27 °\} = \{\frac{{(\sin (90 ° - 68 °))}^{2} + \sin^{2} 68 °}{\cos^{2} 22 ° + \cos^{2} 68 °} + \sin^{2} 63 ° + \cos 63 ° \sin (90 ° - 63 °)\} = \{\frac{\cos^{2} 68 ° + \sin^{2} 68 °}{\cos^{2} 22 ° + \cos^{2} 68 °} + \sin^{2} 63 ° + \cos 63 ° \cos 63 °\} (∵ \sin (90 ° - θ) = \cos θ) = \{\frac{\cos^{2} 68 ° + \sin^{2} 68 °}{{(\cos (90 ° - 68 °))}^{2} + \cos^{2} 68 °} + \sin^{2} 63 ° + \cos^{2} 63 °\} = \{\frac{\cos^{2} 68 ° + \sin^{2} 68 °}{\sin^{2} 68 ° + \cos^{2} 68 °} + \sin^{2} 63 ° + \cos^{2} 63 °\} (∵ \cos (90 ° - θ) = \sin θ) = \{1 + \sin^{2} 63 ° + \cos^{2} 63 °\} = \{1 + 1\} (using the identity : \sin^{2} θ + \cos^{2} θ = 1) = 2 Hence, \{\frac{\sin^{2} 22 ° + \sin^{2} 68 °}{\cos^{2} 22 ° + \cos^{2} 68 °} + \sin^{2} 63 ° + \cos 63 ° \sin 27 °\} = 2 .$

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$\{\frac{\cos 65 °}{\sin 25 °} + \frac{cosec 34 °}{\sec 56 °} - \frac{2 \cos 43 ° cosec 47 °}{\tan 10 ° \tan 40 ° \tan 50 ° \tan 80 °}\} = \{\frac{\cos (90 ° - 25 °)}{\sin 25 °} + \frac{cosec 34 °}{\sec 56 °} - \frac{2 \cos (90 ° - 47 °) cosec 47 °}{\tan 10 ° \tan 40 ° \tan 50 ° \tan 80 °}\} = \{\frac{\sin 25 °}{\sin 25 °} + \frac{cosec 34 °}{\sec 56 °} - \frac{2 \sin 47 ° cosec 47 °}{\tan 10 ° \tan 40 ° \tan 50 ° \tan 80 °}\} (∵ \cos (90 ° - θ) = \sin θ) = \{1 + \frac{cosec (90 ° - 56 °)}{\sec 56 °} - \frac{2 \sin 47 ° cosec 47 °}{\tan 10 ° \tan 40 ° \tan 50 ° \tan 80 °}\} = \{1 + \frac{\sec 56 °}{\sec 56 °} - \frac{2 \sin 47 ° cosec 47 °}{\tan 10 ° \tan 40 ° \tan 50 ° \tan 80 °}\} (∵ cosec (90 ° - θ) = \sec θ) = \{1 + 1 - \frac{2 \sin 47 ° \frac{1}{\sin 47 °}}{\tan 10 ° \tan 40 ° \tan 50 ° \tan 80 °}\} (∵ cosec θ = \frac{1}{\sin θ}) = \{2 - \frac{2}{\tan 10 ° \tan 40 ° \tan 50 ° \tan 80 °}\} = \{2 - \frac{2}{\tan 10 ° \tan 40 ° \tan (90 ° - 40 °) \tan (90 ° - 10 °)}\} = \{2 - \frac{2}{\tan 10 ° \tan 40 ° \cot 40 ° \cot 10 °}\} (∵ \tan (90 ° - θ) = \cot θ) = \{2 - \frac{2}{\tan 10 ° \tan 40 ° \frac{1}{\tan 40 °} \frac{1}{\tan 10 °}}\} (∵ \cot θ = \frac{1}{\tan θ}) = \{2 - \frac{2}{1}\} = 2 - 2 = 0 Hence, \{\frac{\cos 65 °}{\sin 25 °} + \frac{cosec 34 °}{\sec 56 °} - \frac{2 \cos 43 ° cosec 47 °}{\tan 10 ° \tan 40 ° \tan 50 ° \tan 80 °}\} = 0 .$

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$\cot θ \tan (90 ° - θ) - \sec (90 ° - θ) cosec θ + \sin^{2} 65 ° + \sin^{2} 25 ° + \sqrt{3} \tan 5 ° \tan 45 ° \tan 85 ° = \cot θ \cot θ - \sec (90 ° - θ) cosec θ + \sin^{2} 65 ° + \sin^{2} 25 ° + \sqrt{3} \tan 5 ° \tan 45 ° \tan 85 ° (∵ \tan (90 ° - θ) = \cot θ) = \cot^{2} θ - cosec θ cosec θ + \sin^{2} 65 ° + \sin^{2} 25 ° + \sqrt{3} \tan 5 ° \tan 45 ° \tan 85 ° (∵ \sec (90 ° - θ) = cosec θ) = \cot^{2} θ - {cosec}^{2} θ + {(\sin (90 ° - 25 °))}^{2} + \sin^{2} 25 ° + \sqrt{3} \tan 5 ° \tan 45 ° \tan 85 ° = \cot^{2} θ - {cosec}^{2} θ + \cos^{2} 25 ° + \sin^{2} 25 ° + \sqrt{3} \tan 5 ° \tan 45 ° \tan 85 ° (∵ \sin (90 ° - θ) = \cos θ) = - 1 + \cos^{2} 25 ° + \sin^{2} 25 ° + \sqrt{3} \tan 5 ° \tan 45 ° \tan 85 ° (using the identity : {cosec}^{2} θ - \cot^{2} θ = 1) = - 1 + 1 + \sqrt{3} \tan 5 ° \tan 45 ° \tan 85 ° (using the identity : \cos^{2} θ + \sin^{2} θ = 1) = 0 + \sqrt{3} \tan 5 ° \tan 45 ° \tan (90 ° - 5 °) = \sqrt{3} \tan 5 ° \tan 45 ° \cot 5 ° (∵ \tan (90 ° - θ) = \cot θ) = \sqrt{3} \tan 5 ° \tan 45 ° \frac{1}{\tan 5 °} (∵ \cot θ = \frac{1}{\tan θ}) = \sqrt{3} \tan 45 ° = \sqrt{3} (∵ \tan 45 ° = 1) Hence, \cot θ \tan (90 ° - θ) - \sec (90 ° - θ) cosec θ + \sin^{2} 65 ° + \sin^{2} 25 ° + \sqrt{3} \tan 5 ° \tan 45 ° \tan 85 ° = \sqrt{3} .$

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Page No 591:

Answer:

$\sin (50 ° + θ) - \cos (40 ° - θ) + \tan 1 ° \tan 10 ° \tan 20 ° \tan 70 ° \tan 80 ° \tan 89 ° = \sin (50 ° + θ) - \cos (40 ° - θ) + \tan 1 ° \tan 10 ° \tan 20 ° \tan (90 ° - 20 °) \tan (90 ° - 10 °) \tan (90 ° - 1 °) = \sin (50 ° + θ) - \cos (40 ° - θ) + \tan 1 ° \tan 10 ° \tan 20 ° \cot 20 ° \cot 10 ° \cot 1 ° (∵ \tan (90 ° - θ) = \cot θ) = \sin (50 ° + θ) - \cos (40 ° - θ) + \tan 1 ° \tan 10 ° \tan 20 ° \frac{1}{\tan 20 °} \frac{1}{\tan 10 °} \frac{1}{\tan 1 °} (∵ \cot θ = \frac{1}{\tan θ}) = \cos (90 ° - (50 ° + θ)) - \cos (40 ° - θ) + 1 (∵ \cos (90 ° - θ) = \sin θ) = \cos (90 ° - 50 ° - θ) - \cos (40 ° - θ) + 1 = \cos (40 ° - θ) - \cos (40 ° - θ) + 1 = 1 Hence, \sin (50 ° + θ) - \cos (40 ° - θ) + \tan 1 ° \tan 10 ° \tan 20 ° \tan 70 ° \tan 80 ° \tan 89 ° = 1 .$

Page No 593:

Answer:

$\frac{\sec 32 °}{cosec 58 °} = \frac{\sec (90 ° - 58 °)}{cosec 58 °} = \frac{cosec 58 °}{cosec 58 °} (∵ \sec (90 ° - θ) = cosec θ) = 1 Hence, the correct option is (d) .$

Page No 593:

Answer:

$\frac{\cos 12 °}{\sin 78 °} + \frac{\tan 23 °}{\cot 67 °} = \frac{\cos (90 ° - 78 °)}{\sin 78 °} + \frac{\tan 23 °}{\cot 67 °} = \frac{\sin 78 °}{\sin 78 °} + \frac{\tan 23 °}{\cot 67 °} (∵ \cos (90 ° - θ) = \sin θ) = 1 + \frac{\tan (90 ° - 67 °)}{\cot 67 °} = 1 + \frac{\cot 67 °}{\cot 67 °} (∵ \tan (90 ° - θ) = \cot θ) = 1 + 1 = 2 Hence, the correct option is (c) .$

Page No 593:

Answer:

$\tan 10 ° \tan 15 ° \tan 75 ° \tan 80 ° = \tan (90 ° - 80 °) \tan 15 ° \tan (90 ° - 15 °) \tan 80 ° = \cot 80 ° \tan 15 ° \cot 15 ° \tan 80 ° (∵ \tan (90 ° - θ) = \cot θ) = (\frac{1}{\tan 80 °} \tan 80 °) (\tan 15 ° \frac{1}{\tan 15 °}) (∵ \cot θ = \frac{1}{\tan θ}) = 1 Hence, the correct option is (b) .$

Page No 593:

Answer:

$\tan 5 ° \tan 25 ° \tan 30 ° \tan 65 ° \tan 85 ° = \tan (90 ° - 85 °) \tan 25 ° \tan 30 ° \tan (90 ° - 25 °) \tan 85 ° = \cot 85 ° \tan 25 ° \tan 30 ° \cot 25 ° \tan 85 ° (∵ \tan (90 ° - θ) = \cot θ) = (\frac{1}{\tan 85 °} \tan 85 °) (\tan 25 ° \frac{1}{\tan 25 °}) \tan 30 ° (∵ \cot θ = \frac{1}{\tan θ}) = \tan 30 ° = \frac{1}{\sqrt{3}} Hence, the correct option is (c) .$

Page No 593:

Answer:

$\cos 1 ° \cos 2 ° \cos 3 ° . . . . . . \cos 180 ° = \cos 1 ° \cos 2 ° \cos 3 ° . . . . . . \cos 90 ° . . . . . . \cos 180 ° = \cos 1 ° \cos 2 ° \cos 3 ° . . . . . . \times 0 \times . . . . . . \cos 180 ° (∵ \cos 90 ° = 0) = 0 Hence, the correct option is (a) .$

Page No 593:

Answer:

$\sin 43 ° \cos 47 ° + \cos 43 ° \sin 47 ° = \sin (90 ° - 47 °) \cos 47 ° + \cos (90 ° - 47 °) \sin 47 ° = \cos 47 ° \cos 47 ° + \sin 47 ° \sin 47 ° (∵ \sin (90 ° - θ) = \cos θ and \cos (90 ° - θ) = sinθ) = \cos^{2} 47 ° + \sin^{2} 47 ° = 1 (using the identity : \sin^{2} θ + \cos^{2} θ = 1) Hence, the correct option is (b) .$

Page No 593:

Answer:

$\sec 70 ° \sin 20 ° + \cos 20 ° cosec 70 ° = \sec (90 ° - 20 °) \sin 20 ° + \cos 20 ° cosec (90 ° - 20 °) = cosec 20 ° \sin 20 ° + \cos 20 ° \sec 20 ° (∵ \sec (90 ° - θ) = cosec θ and cosec (90 ° - θ) = \sec θ) = \frac{1}{\sin 20 °} \sin 20 ° + \cos 20 ° \frac{1}{\cos 20 °} (∵ cosec θ = \frac{1}{\sin θ} and \sec θ = \frac{1}{\cos θ}) = 1 + 1 = 2 Hence, the correct option is (c) .$

Page No 593:

Answer:

${cosec}^{2} 57 ° - \tan^{2} 33 ° = {(cosec (90 ° - 33 °))}^{2} - \tan^{2} 33 ° = \sec^{2} 33 ° - \tan^{2} 33 ° (∵ cosec (90 ° - θ) = \sec θ) = 1 (using the identity : \sec^{2} θ - \tan^{2} θ = 1) Hence, the correct option is (a) .$

Page No 593:

Answer:

$\sec^{2} 10 ° - \cot^{2} 80 ° = {(\sec (90 ° - 80 °))}^{2} - \cot^{2} 80 ° = {cosec}^{2} 80 ° - \cot^{2} 80 ° (∵ \sec (90 ° - θ) = cosec θ) = 1 (using the identity : {cosec}^{2} θ - \cot^{2} θ = 1) Hence, the correct option is (b) .$

Page No 593:

Answer:

$\frac{2 \sin^{2} 63 ° + 1 + 2 \sin^{2} 27 °}{3 \cos^{2} 17 ° - 2 + 3 \cos^{2} 73 °} = \frac{2 {(\sin (90 ° - 27 °))}^{2} + 1 + 2 \sin^{2} 27 °}{3 {(\cos (90 ° - 73 °))}^{2} - 2 + 3 \cos^{2} 73 °} = \frac{2 \cos^{2} 27 ° + 1 + 2 \sin^{2} 27 °}{3 \sin^{2} 73 ° - 2 + 3 \cos^{2} 73 °} (∵ \sin (90 ° - θ) = \cos θ and \cos (90 ° - θ) = \sin θ) = \frac{2 (\cos^{2} 27 ° + \sin^{2} 27 °) + 1}{3 (\sin^{2} 73 ° + \cos^{2} 73 °) - 2} = \frac{2 (1) + 1}{3 (1) - 2} (using the identity : \sin^{2} θ + \cos^{2} θ = 1) = \frac{2 + 1}{3 - 2} = 3 Hence, the correct option is (d) .$

Page No 594:

Answer:

$\sin 38 ° - \cos 52 ° = \sin (90 ° - 52 °) - \cos 52 ° = \cos 52 ° - \cos 52 ° (∵ \sin (90 ° - θ) = \cos θ) = 0 Hence, the correct option is (a) .$

Page No 594:

Answer:

$\frac{(\sin^{2} 23 ° + \sin^{2} 67 °)}{(\cos^{2} 13 ° + \cos^{2} 77 °)} + \sin^{2} 59 ° + \cos 59 ° \sin 31 ° = [\frac{\sin^{2} 23 ° + {(\sin (90 ° - 23 °))}^{2}}{\cos^{2} 13 ° + {(\cos (90 ° - 13 °))}^{2}}] + \sin^{2} 59 ° + \cos 59 ° \sin (90 ° - 59 °) = [\frac{\sin^{2} 23 ° + \cos^{2} 23 °}{\cos^{2} 13 ° + \sin^{2} 13 °}] + \sin^{2} 59 ° + \cos 59 ° \cos 59 ° (∵ \sin (90 ° - θ) = \cos θ and \cos (90 ° - θ) = \sin θ) = [\frac{1}{1}] + \sin^{2} 59 ° + \cos^{2} 59 ° (using the identity : \sin^{2} θ + \cos^{2} θ = 1) = 1 + 1 = 2 Hence, the correct option is (b) .$

Disclaimer: There must be $\sin 31 °$ instead of $\sin 21 °$ .

Page No 594:

Answer:

$(\frac{2 \tan^{2} 30 ° \sec^{2} 52 ° \sin^{2} 38 °}{{cosec}^{2} 70 ° - \tan^{2} 20 °}) = (\frac{2 \tan^{2} 30 ° {(\sec (90 ° - 38 °))}^{2} \sin^{2} 38 °}{{(cosec (90 ° - 20 °))}^{2} - \tan^{2} 20 °}) = (\frac{2 \tan^{2} 30 ° {cosec}^{2} 38 ° \sin^{2} 38 °}{\sec^{2} 20 ° - \tan^{2} 20 °}) (∵ \sec (90 ° - θ) = cosec θ and cosec (90 ° - θ) = \sec θ) = (\frac{2 \tan^{2} 30 ° {cosec}^{2} 38 ° \sin^{2} 38 °}{1}) (using the identity : \sec^{2} θ - \tan^{2} θ = 1) = 2 \tan^{2} 30 ° \frac{1}{\sin^{2} 38 °} \sin^{2} 38 ° (∵ cosec θ = \frac{1}{\sin θ}) = 2 \tan^{2} 30 ° = 2 {(\frac{1}{\sqrt{3}})}^{2} (∵ \tan 30 ° = \frac{1}{\sqrt{3}}) = \frac{2}{3} Hence, the correct option is (c) .$

Page No 594:

Answer:

$(\frac{\cos 38 ° cosec 52 °}{\tan 18 ° \tan 35 ° \tan 60 ° \tan 72 ° \tan 55 °}) = (\frac{\cos 38 ° cosec 52 °}{\tan (90 ° - 72 °) \tan (90 ° - 55 °) \tan 60 ° \tan 72 ° \tan 55 °}) = (\frac{\cos 38 ° cosec (90 ° - 38 °)}{\cot 72 ° \cot 55 ° \tan 60 ° \tan 72 ° \tan 55 °}) (∵ \tan (90 ° - θ) = \cot θ) = (\frac{\cos 38 ° \sec 38 °}{\cot 72 ° \cot 55 ° \tan 60 ° \tan 72 ° \tan 55 °}) (∵ cosec (90 ° - θ) = \sec θ) = (\frac{\cos 38 ° \frac{1}{\cos 38 °}}{\frac{1}{\tan 72 °} \frac{1}{\tan 55 °} \tan 60 ° \tan 72 ° \tan 55 °}) (∵ \sec θ = \frac{1}{\cos θ} and \cot θ = \frac{1}{\tan θ}) = (\frac{1}{\tan 60 °}) = \frac{1}{\sqrt{3}} (∵ \tan 60 ° = \sqrt{3}) Hence, the correct option is (b) .$

Page No 594:

Answer:

$\sin (60 ° + θ) - \cos (30 ° - θ) = \cos (90 ° - (60 ° + θ)) - \cos (30 ° - θ) (∵ \sin θ = \cos (90 ° - θ)) = \cos (90 ° - 60 ° - θ) - \cos (30 ° - θ) = \cos (30 ° - θ) - \cos (30 ° - θ) = 0 Hence, the correct option is (c) .$

Page No 594:

Answer:

Given: sinA = cosB

$\sin A = \cos B \Rightarrow \cos (90 ° - A) = \cos B (∵ \sin θ = \cos (90 ° - θ)) \Rightarrow 90 ° - A = B \Rightarrow 90 ° = B + A \Rightarrow A + B = 90 ° Hence, the correct option is (d) .$

Page No 594:

Answer:

Given: cos(α + β) = 0

$As we know that, \cos 90 ° = 0 Since, \cos (α + β) = 0 \Rightarrow α + β = 90 ° \Rightarrow α = 90 ° - β . . . (1) Now, \sin (α - β) = \sin (90 ° - β - β) = \sin (90 ° - 2 β) = \cos 2 β (∵ \sin (90 ° - θ) = \cos θ) Hence, the correct option is (b) .$

Page No 594:

Answer:

$\sin (45 ° + θ) - \cos (45 ° - θ) = \cos (90 ° - (45 ° + θ)) - \cos (45 ° - θ) (∵ \sin θ = \cos (90 ° - θ)) = \cos (90 ° - 45 ° - θ) - \cos (45 ° - θ) = \cos (45 ° - θ) - \cos (45 ° - θ) = 0 Hence, the correct option is (a) .$

Page No 594:

Answer:

Given: sec4A = cosec(A – 10°)

$\sec 4 A = cosec (A - 10 °) \Rightarrow cosec (90 ° - 4 A) = cosec (A - 10 °) (∵ \sec θ = cosec (90 ° - θ)) \Rightarrow 90 ° - 4 A = A - 10 ° \Rightarrow 90 ° + 10 ° = A + 4 A \Rightarrow 5 A = 100 ° \Rightarrow A = \frac{100 °}{5} \Rightarrow A = 20 ° Hence, the correct option is (a) .$

Page No 594:

Answer:

Page No 594:

Answer:

(i) cot34° – tan56° = .......

$\cot 34 ° - \tan 56 ° = \cot (90 ° - 56 °) - \tan 56 ° = \tan 56 ° - \tan 56 ° (∵ \cot (90 ° - θ) = \tan θ) = 0 Hence, \cot 34 ° - \tan 56 ° = 0 .$

(ii) cosec31° – sec59° = .......

$cosec 31 ° - \sec 59 ° = cosec (90 ° - 59 °) - \sec 59 ° = \sec 59 ° - \sec 59 ° (∵ cosec (90 ° - θ) = \sec θ) = 0 Hence, cosec 31 ° - \sec 59 ° = 0 .$

(iii) cos²67° + cos²23° = ..........

$\cos^{2} 67 ° + \cos^{2} 23 ° = {(\cos (90 ° - 23 °))}^{2} + \cos^{2} 23 ° (∵ \cos (90 ° - θ) = \sin θ) = \sin^{2} 23 ° + \cos^{2} 23 ° (∵ \sin^{2} θ + \cos^{2} θ = 1) = 1 Hence, \cos^{2} 67 ° + \cos^{2} 23 ° = 1 .$

(iv) cosec²54° – tan²36° = .........

${cosec}^{2} 54 ° - \tan^{2} 36 ° = {(cosec (90 ° - 36 °))}^{2} - \tan^{2} 36 ° (∵ cosec (90 ° - θ) = \sec θ) = \sec^{2} 36 ° - \tan^{2} 36 ° (∵ \sec^{2} θ - \tan^{2} θ = 1) = 1 Hence, {cosec}^{2} 54 ° - \tan^{2} 36 ° = 1 .$

(v) sec²40° – cot²50° = ..........

$\sec^{2} 40 ° - \cot^{2} 50 ° = {(\sec (90 ° - 50 °))}^{2} - \cot^{2} 50 ° (∵ \sec (90 ° - θ) = cosec θ) = {cosec}^{2} 50 ° - \cot^{2} 50 ° (∵ {cosec}^{2} θ - \cot^{2} θ = 1) = 1 Hence, \sec^{2} 40 ° - \cot^{2} 50 ° = 1 .$

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