CBSE Class 10 Introduction to Trigonometry Sure Shot Questions Set F

Question. Solve : \( \frac{2}{3} \text{cosec}^2 58^\circ - \frac{2}{3} \cot 58^\circ \tan 32^\circ - \frac{5}{3} \tan 13^\circ \tan 37^\circ \tan 45^\circ \tan 53^\circ \tan 77^\circ \).
Answer: Consider,
\( \frac{2}{3} \text{cosec}^2 58^\circ - \frac{2}{3} \cot 58^\circ \tan 32^\circ - \frac{5}{3} \tan 13^\circ \tan 37^\circ \tan 45^\circ \tan 53^\circ \tan 77^\circ \)
\( = \frac{2}{3} \text{cosec}^2 58^\circ - \frac{2}{3} \cot 58^\circ \tan (90^\circ - 58^\circ) - \frac{5}{3} \tan 13^\circ \tan 37^\circ \tan 45^\circ \tan (90^\circ - 37^\circ) \tan (90^\circ - 13^\circ) \)
\( = \frac{2}{3} \text{cosec}^2 58^\circ - \frac{2}{3} \cot 58^\circ \cot 58^\circ - \frac{5}{3} \tan 13^\circ \tan 37^\circ \tan 45^\circ \cot 37^\circ \cot 13^\circ \)
\( = \frac{2}{3} (\text{cosec}^2 58^\circ - \cot^2 58^\circ) - \frac{5}{3} (\tan 13^\circ \cot 13^\circ)(\tan 37^\circ \cot 37^\circ) \tan 45^\circ \)
\( = \frac{2}{3}(1) - \frac{5}{3}(1)(1)(1) = \frac{2}{3} - \frac{5}{3} = -1 \).

Question. Evaluate : 
\[ \left( \frac{3 \sin 43^\circ}{\cos 47^\circ} \right)^2 - \frac{\cos 37^\circ \text{cosec } 53^\circ}{\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ} \]
Answer: Sol.
\[ \left( \frac{3 \sin 43^\circ}{\cos 47^\circ} \right)^2 - \frac{\cos 37^\circ \text{cosec } 53^\circ}{\tan 5^\circ \tan 25^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ} \]
= \( \left\{ \frac{3 \cos(90^\circ - 43^\circ)}{\cos 47^\circ} \right\}^2 - \frac{\sin(90^\circ - 37^\circ) \text{cosec } 53^\circ}{\cot(90^\circ - 5^\circ) \cot(90^\circ - 25^\circ) \tan 45^\circ \tan 65^\circ \tan 85^\circ} \)
= \( \left\{ \frac{3 \cos 47^\circ}{\cos 47^\circ} \right\}^2 - \frac{\sin 53^\circ \text{cosec } 53^\circ}{\cot 85^\circ \cot 65^\circ \tan 45^\circ \tan 65^\circ \tan 85^\circ} \)
[\(\because \cos(90^\circ - \theta) = \sin \theta, \sin(90^\circ - \theta) = \cos \theta, \cot(90^\circ - \theta) = \tan \theta\)]
= \( (3)^2 - \frac{\sin 53^\circ \times \frac{1}{\sin 53^\circ}}{\frac{1}{\tan 85^\circ} \times \frac{1}{\tan 65^\circ} \times \tan 45^\circ \times \tan 65^\circ \times \tan 85^\circ} \)
[\(\because \sin \theta = \frac{1}{\text{cosec } \theta}, \tan \theta = \frac{1}{\cot \theta}\)]
= \( 9 - \frac{1}{\tan 45^\circ} \)
= \( 9 - 1 \) [\(\because \tan 45^\circ = 1\)]
= 8 Ans.

Question. Solve :
\( -\tan \theta \cot (90^\circ - \theta) + \frac{\sec \theta \text{cosec } (90^\circ - \theta) + \sin^2 35^\circ + \sin^2 55^\circ}{\tan 10^\circ \tan 20^\circ \tan 30^\circ \tan 70^\circ \tan 80^\circ} \)
Answer: Sol. We have, \( -\tan \theta \cot (90^\circ - \theta) + \frac{\sec \theta \text{cosec } (90^\circ - \theta) + \sin^2 35^\circ + \sin^2 55^\circ}{\tan 10^\circ \tan 20^\circ \tan 30^\circ \tan 70^\circ \tan 80^\circ} \)
= \( -\tan \theta \tan \theta + \frac{\sec \theta \sec \theta + \sin^2 35^\circ + \sin^2 (90^\circ - 35^\circ)}{\tan 10^\circ \tan 20^\circ \tan 30^\circ \tan (90^\circ - 20^\circ) \tan (90^\circ - 10^\circ)} \)
= \( \frac{1 + \sin^2 35^\circ + \cos^2 35^\circ}{\tan 10^\circ \tan 20^\circ \tan 30^\circ \cot 20^\circ \cot 10^\circ} \)
= \( \frac{1 + 1}{\tan 30^\circ} = 2\sqrt{3} \). Ans.

Question. If \(\sin \theta + \sin^2 \theta = 1\), find the value of \(\cos^{12} \theta + 3 \cos^{10} \theta + 3 \cos^8 \theta + \cos^6 \theta + 2 \cos^4 \theta + 2 \cos^2 \theta - 2\)
Answer: Sol. We have,
\(\sin \theta + \sin^2 \theta = 1\)
\(\Rightarrow \sin \theta = 1 - \sin^2 \theta\)
\(\Rightarrow \sin \theta = \cos^2 \theta\)
Now,
\(\cos^{12} \theta + 3 \cos^{10} \theta + 3 \cos^8 \theta + \cos^6 \theta + 2 \cos^4 \theta + 2 \cos^2 \theta - 2\)
= \((\cos^{12} \theta + 3 \cos^{10} \theta + 3 \cos^8 \theta + \cos^6 \theta) + 2(\cos^4 \theta + \cos^2 \theta - 1)\)
= \((\cos^4 \theta + \cos^2 \theta)^3 + 2(\cos^4 \theta + \cos^2 \theta - 1)\)
= \((\sin^2 \theta + \cos^2 \theta)^3 + 2(\sin^2 \theta + \cos^2 \theta - 1)\)
[\(\because \cos^2 \theta = \sin \theta \therefore \cos^4 \theta = \sin^2 \theta\)]
= \(1 + 2(1 - 1) = 1\) Ans.

Question. If \(\tan^2 \theta = 1 - a^2\), prove that \(\sec \theta + \tan^3 \theta \text{cosec } \theta = (2 - a^2)^{3/2}\)
Answer: Sol. We have,
\(\sec \theta + \tan^3 \theta \text{cosec } \theta\)
= \(\sec \theta \left\{ \frac{\sec \theta + \tan^3 \theta \text{cosec } \theta}{\sec \theta} \right\}\)
[Multiplying and dividing by \(\sec \theta\)]
= \(\sec \theta \left\{ 1 + \tan^3 \theta \cdot \frac{\cos \theta}{\sin \theta} \right\}\)
= \(\sec \theta \{1 + \tan^3 \theta \times \cot \theta\}\)
= \(\sqrt{1 + \tan^2 \theta} \{1 + \tan^2 \theta\}\)
= \(\{1 + \tan^2 \theta\}^{3/2}\)
= \(\{1 + (1 - a^2)\}^{3/2}\) [\(\because \tan^2 \theta = 1 - a^2\); Given]
= \((2 - a^2)^{3/2}\) Hence Proved.

Question. If \(x = a \cos^3 \theta, y = b \sin^3 \theta\), prove that \(\left( \frac{x}{a} \right)^{2/3} + \left( \frac{y}{b} \right)^{2/3} = 1\).
Answer: Sol. Given,
\(x = a \cos^3 \theta\) and \(y = b \sin^3 \theta\)
L.H.S. = \(\left( \frac{x}{a} \right)^{2/3} + \left( \frac{y}{b} \right)^{2/3}\)
= \(\left( \frac{a \cos^3 \theta}{a} \right)^{2/3} + \left( \frac{b \sin^3 \theta}{b} \right)^{2/3}\)
= \((\cos \theta)^{3 \times 2/3} + (\sin \theta)^{3 \times 2/3}\)
= \(\cos^2 \theta + \sin^2 \theta = 1 = \text{R.H.S.}\)
Hence Proved.

Question. Prove that : (1 + \cot A - \text{cosec } A) (1 + \tan A + \sec A) = 2.
Answer: Sol. L.H.S. = \((1 + \cot A - \text{cosec } A) (1 + \tan A + \sec A)\)
= \(\left( 1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A} \right) \left( 1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A} \right)\)
= \(\left( \frac{\sin A + \cos A - 1}{\sin A} \right) \left( \frac{\cos A + \sin A + 1}{\cos A} \right)\)
= \(\frac{(\sin A + \cos A)^2 - 1}{\sin A \cdot \cos A}\) [\(\because (a + b) (a - b) = a^2 - b^2\)]
= \(\frac{\sin^2 A + \cos^2 A + 2 \cdot \sin A \cdot \cos A - 1}{\sin A \cdot \cos A}\)
= \(\frac{1 + 2 \sin A \cdot \cos A - 1}{\sin A \cdot \cos A} = \frac{2 \sin A \cdot \cos A}{\sin A \cdot \cos A}\)
= 2 (R.H.S.) Hence Proved.

Question. Prove that \(\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A\) 
Answer: Sol. Consider, L.H.S. = \(\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}\)
= \(\frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A) \cos A}\)
= \(\frac{\cos^2 A + \sin^2 A + 1 + 2 \sin A}{(1 + \sin A) \cos A}\)
= \(\frac{1 + 1 + 2 \sin A}{(1 + \sin A) \cos A}\) [\(\because \cos^2 A + \sin^2 A = 1\)]
= \(\frac{2 + 2 \sin A}{(1 + \sin A) \cos A}\)
= \(\frac{2 (1 + \sin A)}{(1 + \sin A) \cos A} = \frac{2}{\cos A}\)
= 2 sec A = R.H.S. Hence Proved.

Long Answer Type Questions

Question. Prove that : \(\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{ cosec } \theta\)
Answer: Sol. L.H.S.
\(= \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta}\)
\(= \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}\)
\(= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}\)
\(= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}\)
\(= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}\)
\(= \frac{1}{\sin \theta - \cos \theta} \left[ \frac{\sin^2 \theta}{\cos \theta} - \frac{\cos^2 \theta}{\sin \theta} \right]\)
\(= \frac{1}{\sin \theta - \cos \theta} \left[ \frac{\sin^3 \theta - \cos^3 \theta}{\cos \theta \cdot \sin \theta} \right]\)
\(= \frac{[\sin \theta - \cos \theta][\sin^2 \theta + \cos^2 \theta + \sin \theta \cdot \cos \theta]}{(\sin \theta - \cos \theta) \cdot (\cos \theta \sin \theta)}\)
[\(\because a^3 - b^3 = (a - b) (a^2 + b^2 + ab)\)]
\(= \frac{\sin^2 \theta + \cos^2 \theta + \sin \theta \cdot \cos \theta}{(\cos \theta \times \sin \theta)}\)
\(= \frac{1 + \sin \theta \cdot \cos \theta}{\cos \theta \times \sin \theta}\) [\(\because \sin^2 \theta + \cos^2 \theta = 1\)]
\(= \frac{1}{\cos \theta \cdot \sin \theta} + \frac{\sin \theta \cdot \cos \theta}{\sin \theta \cdot \cos \theta}\)
\(= 1 + \sec \theta \cdot \text{cosec } \theta = \text{R.H.S.}\)
\(\left( \because \frac{1}{\cos \theta} = \sec \theta, \frac{1}{\sin \theta} = \text{cosec } \theta \right)\)
Hence Proved.

Question. Prove that : [2019 O.D.] \(\frac{\sin \theta}{\cot \theta + \text{cosec } \theta} = 2 + \frac{\sin \theta}{\cot \theta - \text{cosec } \theta}\)
Answer: Sol. L.H.S. = \(\frac{\sin \theta}{\cot \theta + \text{cosec } \theta}\)
\(= \frac{\sin \theta}{\frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta}}\)
\(= \frac{\sin \theta}{\frac{\cos \theta + 1}{\sin \theta}}\)
\(= \frac{\sin^2 \theta}{\cos \theta + 1}\)
\(= \frac{\sin^2 \theta}{1 + \cos \theta} \times \frac{(1 - \cos \theta)}{(1 - \cos \theta)}\)
\(= \frac{\sin^2 \theta (1 - \cos \theta)}{1 - \cos^2 \theta}\)
\(= \frac{\sin^2 \theta (1 - \cos \theta)}{\sin^2 \theta}\)
\(= 1 - \cos \theta\) ...(i)
R.H.S. = \(2 + \frac{\sin \theta}{\cot \theta - \text{cosec } \theta}\)
\(= 2 + \frac{\sin \theta}{\frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}}\)
\(= 2 + \frac{\sin^2 \theta}{\cos \theta - 1}\)
\(= 2 - \frac{\sin^2 \theta}{(1 - \cos \theta)}\)
\(= 2 - \frac{\sin^2 \theta \times (1 + \cos \theta)}{(1 - \cos \theta) \times (1 + \cos \theta)}\)
\(= 2 - \frac{\sin^2 \theta (1 + \cos \theta)}{1 - \cos^2 \theta}\)
\(= 2 - \frac{\sin^2 \theta (1 + \cos \theta)}{\sin^2 \theta}\)
\(= 2 - (1 + \cos \theta)\)
\(= 1 - \cos \theta\) ...(ii)
From equation (i) and (ii), we get
L.H.S. = R.H.S. Hence Proved.

Question. Prove that \(\frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} = \frac{1}{\sec A - \tan A}\) 
Answer: Sol. L.H.S. = \(\frac{\sin A - \cos A + 1}{\sin A + \cos A - 1}\)
Dividing the numerator and denominator by \(\cos A\)
\(= \frac{\tan A - 1 + \sec A}{\tan A + 1 - \sec A}\)
\(= \frac{(\tan A + \sec A) - 1}{(\tan A - \sec A) + 1}\)
\(= \frac{(\tan A + \sec A) - (\sec^2 A - \tan^2 A)}{\tan A - \sec A + 1}\) [\(\because \sec^2 A - \tan^2 A = 1\)]
\(= \frac{(\tan A + \sec A) [1 - (\sec A - \tan A)]}{(\tan A - \sec A + 1)}\)
\(= \frac{(\tan A + \sec A) [1 - \sec A + \tan A]}{(\tan A - \sec A + 1)}\)
\(= \tan A + \sec A\)
\(= (\tan A + \sec A) \times \frac{(\sec A - \tan A)}{(\sec A - \tan A)}\)
\(= \frac{\sec^2 A - \tan^2 A}{\sec A - \tan A} = \frac{1}{\sec A - \tan A} = \text{R.H.S.}\)
Hence Proved.

Question. Prove that : [2019 Delhi] \(\frac{\tan^2 A}{\tan^2 A - 1} + \frac{\text{cosec}^2 A}{\sec^2 A - \text{cosec}^2 A} = \frac{1}{1 - 2 \cos^2 A}\)
Answer: Sol. L.H.S. = \(\frac{\tan^2 A}{\tan^2 A - 1} + \frac{\text{cosec}^2 A}{\sec^2 A - \text{cosec}^2 A}\)
\(= \frac{\frac{\sin^2 A}{\cos^2 A}}{\frac{\sin^2 A}{\cos^2 A} - 1} + \frac{\frac{1}{\sin^2 A}}{\frac{1}{\cos^2 A} - \frac{1}{\sin^2 A}}\)
\(= \frac{\frac{\sin^2 A}{\cos^2 A}}{\frac{\sin^2 A - \cos^2 A}{\cos^2 A}} + \frac{\frac{1}{\sin^2 A}}{\frac{\sin^2 A - \cos^2 A}{\cos^2 A \sin^2 A}}\)
\(= \frac{\sin^2 A}{\sin^2 A - \cos^2 A} + \frac{1}{\sin^2 A} \cdot \frac{\cos^2 A \sin^2 A}{\sin^2 A - \cos^2 A}\)
\(= \frac{\sin^2 A}{\sin^2 A - \cos^2 A} + \frac{\cos^2 A}{\sin^2 A - \cos^2 A}\)
\(= \frac{\sin^2 A + \cos^2 A}{\sin^2 A - \cos^2 A}\)
\(= \frac{1}{(1 - \cos^2 A) - \cos^2 A}\) [\(\because \sin^2 A = 1 - \cos^2 A\)]
\(= \frac{1}{1 - 2 \cos^2 A} = \text{R.H.S.}\)
Hence Proved.

Question. Prove that : \(\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} = \tan A\) 
Answer: Sol. L.H.S. = \(\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A}\)
\(= \frac{\sin A (1 - 2 \sin^2 A)}{\cos A (2 \cos^2 A - 1)}\)
\(= \frac{\sin A (\sin^2 A + \cos^2 A - 2 \sin^2 A)}{\cos A (2 \cos^2 A - (\sin^2 A + \cos^2 A))}\) [\(\because \sin^2 A + \cos^2 A = 1\)]
\(= \frac{\sin A (\cos^2 A - \sin^2 A)}{\cos A (2 \cos^2 A - \sin^2 A - \cos^2 A)}\)
\(= \frac{\sin A (\cos^2 A - \sin^2 A)}{\cos A (\cos^2 A - \sin^2 A)}\)
\(= \frac{\sin A}{\cos A} \times 1\)
\(= \tan A\).
L.H.S. = R.H.S.
Hence Proved.

Question. Prove that : \(\frac{\sec A - 1}{\sec A + 1} = \left( \frac{\sin A}{1 + \cos A} \right)^2 = (\cot A - \text{cosec } A)^2\)
Answer: Sol. L.H.S. = \(\frac{\sec A - 1}{\sec A + 1}\)
\(= \frac{\frac{1}{\cos A} - 1}{\frac{1}{\cos A} + 1}\)
\(= \frac{\frac{1 - \cos A}{\cos A}}{\frac{1 + \cos A}{\cos A}} = \frac{1 - \cos A}{1 + \cos A}\)
\(= \frac{(1 - \cos A)(1 + \cos A)}{(1 + \cos A)(1 + \cos A)}\)
\(= \frac{1 - \cos^2 A}{(1 + \cos A)^2}\)
\(= \frac{\sin^2 A}{(1 + \cos A)^2}\) [\(\because 1 - \cos^2 A = \sin^2 A\)]
\(= \left( \frac{\sin A}{1 + \cos A} \right)^2\)
And \(\left( \frac{\sin A}{1 + \cos A} \right)^2 = \left[ \frac{\sin A}{1 + \cos A} \times \frac{(1 - \cos A)}{(1 - \cos A)} \right]^2\)
\(= \left[ \frac{\sin A(1 - \cos A)}{1 - \cos^2 A} \right]^2\)
\(= \left[ \frac{\sin A(1 - \cos A)}{\sin^2 A} \right]^2\)
\(= \left[ \frac{1 - \cos A}{\sin A} \right]^2\)
\(= \left[ \frac{1}{\sin A} - \frac{\cos A}{\sin A} \right]^2\)
\(= (\text{cosec } A - \cot A)^2\)
\(= (-1)^2 (\cot A - \text{cosec } A)^2\)
\(= (\cot A - \text{cosec } A)^2 = \text{R.H.S.}\)
Hence Proved.

Question. If \(m = \tan \theta + \sin \theta\) and \(n = \tan \theta - \sin \theta\), then \((m^2 - n^2)^2 = 16mn\). Is this statement correct ?
Answer: Sol. We have,
\(m^2 = (\tan \theta + \sin \theta)^2 = \tan^2 \theta + \sin^2 \theta + 2 \tan \theta \cdot \sin \theta\)
\(n^2 = (\tan \theta - \sin \theta)^2 = \tan^2 \theta + \sin^2 \theta - 2 \tan \theta \cdot \sin \theta\)
Thus, \(m^2 - n^2 = 4 \tan \theta \cdot \sin \theta\)
L.H.S. = \((m^2 - n^2)^2 = 16 \tan^2 \theta \cdot \sin^2 \theta = 16 \frac{\sin^4 \theta}{\cos^2 \theta}\)
R.H.S. = \(16 mn\)
\(= 16(\tan \theta + \sin \theta) (\tan \theta - \sin \theta)\)
\(= 16(\tan^2 \theta - \sin^2 \theta)\)
\(= 16 \left( \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta \right)\)
\(= 16 \sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right)\)
\(= 16 \sin^2 \theta (\sec^2 \theta - 1)\)
\(= 16 \sin^2 \theta \cdot \tan^2 \theta\)
L.H.S. = R.H.S. Hence correct.

Question. If \( a = \sin \theta + \cos \theta \); \( b = \sin^3 \theta + \cos^3 \theta \), then show that \( (3a - 2b) = a^3 \).
Answer: Sol. Given,
\( a = \sin \theta + \cos \theta \) ...(i)
\( b = \sin^3 \theta + \cos^3 \theta \) ...(ii)
Here \( (\sin \theta + \cos \theta)^2 = a^2 \)
\( \Rightarrow \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = a^2 \)
\( \Rightarrow \sin \theta \cos \theta = \frac{a^2 - 1}{2} \) ...(iii)
\( [\because \sin^2 \theta + \cos^2 \theta = 1] \)
Also, \( (\sin \theta + \cos \theta)^3 = a^3 \)
\( \Rightarrow \sin^3 \theta + \cos^3 \theta + 3 \sin \theta \cos \theta (\sin \theta + \cos \theta) = a^3 \)
\( \Rightarrow \sin^3 \theta + \cos^3 \theta + 3 \left( \frac{a^2 - 1}{2} \right) a = a^3 \)
[From (i) and (iii)]
\( \Rightarrow b + \frac{3a}{2} (a^2 - 1) = a^3 \)
\( \Rightarrow 2b + 3a(a^2 - 1) = 2a^3 \) [From (ii)]
\( \Rightarrow 2b + 3a^3 - 3a = 2a^3 \)
\( \Rightarrow (3a - 2b) = a^3 \) Hence Proved.

Question. If \( a \cos \theta - b \sin \theta = c \), prove that \( (a \sin \theta + b \cos \theta) = \sqrt{a^2 + b^2 - c^2} \).
Answer: Sol. Given, \( a \cos \theta - b \sin \theta = c \) ...(i)
Now, \( a^2 + b^2 - c^2 = a^2 + b^2 - (a \cos \theta - b \sin \theta)^2 \)
[From (i)]
\( = a^2 + b^2 - (a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \sin \theta \cos \theta) \)
\( = a^2 + b^2 - [a^2 (1 - \sin^2 \theta) + b^2 (1 - \cos^2 \theta) - 2ab \sin \theta \cos \theta] \)
\( = a^2 + b^2 - [a^2 - a^2 \sin^2 \theta + b^2 - b^2 \cos^2 \theta - 2ab \sin \theta \cos \theta] \)
\( = a^2 + b^2 - a^2 + a^2 \sin^2 \theta - b^2 + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta \)
\( = a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \sin \theta \cos \theta \)
\( = (a \sin \theta + b \cos \theta)^2 \)
Thus, \( a^2 + b^2 - c^2 = (a \sin \theta + b \cos \theta)^2 \)
or \( \sqrt{a^2 + b^2 - c^2} = (a \sin \theta + b \cos \theta) \)
\( \Rightarrow (a \sin \theta + b \cos \theta) = \sqrt{a^2 + b^2 - c^2} \).
Hence Proved.

Question. If \( \sec \theta + \tan \theta = p \), prove that \( \sin \theta = \frac{p^2 - 1}{p^2 + 1} \).
Answer: Sol. Given, \( \sec \theta + \tan \theta = p \)
Now, R.H.S. \( = \frac{p^2 - 1}{p^2 + 1} \)
\( = \frac{(\sec \theta + \tan \theta)^2 - 1}{(\sec \theta + \tan \theta)^2 + 1} \)
\( = \frac{\sec^2 \theta + \tan^2 \theta + 2 \sec \theta \tan \theta - 1}{\sec^2 \theta + \tan^2 \theta + 2 \sec \theta \tan \theta + 1} \) [\( \because (a + b)^2 = a^2 + b^2 + 2ab \)]
\( = \frac{(\sec^2 \theta - 1) + \tan^2 \theta + 2 \sec \theta \tan \theta}{\sec^2 \theta + (1 + \tan^2 \theta) + 2 \sec \theta \tan \theta} \)
\( = \frac{\tan^2 \theta + \tan^2 \theta + 2 \sec \theta \tan \theta}{\sec^2 \theta + \sec^2 \theta + 2 \sec \theta \tan \theta} \) [\( \because \sec^2 \theta - 1 = \tan^2 \theta \); \( \sec^2 \theta = 1 + \tan^2 \theta \)]
\( = \frac{2 \tan^2 \theta + 2 \sec \theta \tan \theta}{2 \sec^2 \theta + 2 \sec \theta \tan \theta} \)
\( = \frac{2 \tan \theta (\tan \theta + \sec \theta)}{2 \sec \theta (\sec \theta + \tan \theta)} = \frac{\tan \theta}{\sec \theta} \)
\( = \frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta}} \)
\( = \sin \theta = \text{L.H.S.} \) Hence Proved.

Question. If \( \text{cosec } \theta - \sin \theta = m \) and \( \sec \theta - \cos \theta = n \), prove that \( (m^2n)^{2/3} + (mn^2)^{2/3} = 1 \).
Answer: Sol. Given,
\( \text{cosec } \theta - \sin \theta = m \)
\( \Rightarrow \frac{1}{\sin \theta} - \sin \theta = m \)
\( \Rightarrow \frac{1 - \sin^2 \theta}{\sin \theta} = m \)
\( \Rightarrow \frac{\cos^2 \theta}{\sin \theta} = m \)
Also, \( \sec \theta - \cos \theta = n \)
\( \Rightarrow \frac{1}{\cos \theta} - \cos \theta = n \)
\( \Rightarrow \frac{1 - \cos^2 \theta}{\cos \theta} = n \)
\( \Rightarrow \frac{\sin^2 \theta}{\cos \theta} = n \)
Now, consider,
L.H.S. \( = (m^2n)^{2/3} + (mn^2)^{2/3} \)
\( = \left\{ \left( \frac{\cos^2 \theta}{\sin \theta} \right)^2 \left( \frac{\sin^2 \theta}{\cos \theta} \right) \right\}^{2/3} + \left\{ \left( \frac{\cos^2 \theta}{\sin \theta} \right) \left( \frac{\sin^2 \theta}{\cos \theta} \right)^2 \right\}^{2/3} \)
\( = \left( \frac{\cos^4 \theta \cdot \sin^2 \theta}{\sin^2 \theta \cdot \cos \theta} \right)^{2/3} + \left( \frac{\cos^2 \theta \cdot \sin^4 \theta}{\sin \theta \cdot \cos^2 \theta} \right)^{2/3} \)
\( = (\cos^3 \theta)^{2/3} + (\sin^3 \theta)^{2/3} \)
\( = \cos^2 \theta + \sin^2 \theta \)
\( = 1 = \text{R.H.S.} \) Hence Proved.

Question. Prove that \( \frac{1}{1 + \sin^2 \theta} + \frac{1}{1 + \cos^2 \theta} + \frac{1}{1 + \sec^2 \theta} + \frac{1}{1 + \text{cosec}^2 \theta} = 2 \) 
Answer: Sol. Taking from LHS,
LHS \( = \frac{1}{1 + \sin^2 \theta} + \frac{1}{1 + \cos^2 \theta} + \frac{1}{1 + \sec^2 \theta} + \frac{1}{1 + \text{cosec}^2 \theta} \)
\( = \frac{1}{1 + \sin^2 \theta} + \frac{1}{1 + \text{cosec}^2 \theta} + \frac{1}{1 + \cos^2 \theta} + \frac{1}{1 + \sec^2 \theta} \) [Re-arranging]
\( = \frac{1}{1 + \sin^2 \theta} + \frac{1}{1 + \left( \frac{1}{\sin^2 \theta} \right)} + \frac{1}{1 + \cos^2 \theta} + \frac{1}{1 + \left( \frac{1}{\cos^2 \theta} \right)} \)
\( = \frac{1}{1 + \sin^2 \theta} + \frac{\sin^2 \theta}{1 + \sin^2 \theta} + \frac{1}{1 + \cos^2 \theta} + \frac{\cos^2 \theta}{1 + \cos^2 \theta} \)
\( = \frac{1 + \sin^2 \theta}{1 + \sin^2 \theta} + \frac{1 + \cos^2 \theta}{1 + \cos^2 \theta} \)
\( = 1 + 1 = 2 \)
\( = \text{RHS} \). Hence, proved!

Question. If \( \sec \theta = x + \frac{1}{4x} \), prove that \( \sec \theta + \tan \theta = 2x \) or \( \frac{1}{2x} \).
Answer: Sol. Given, \( \sec \theta = x + \frac{1}{4x} \)
On squaring both sides, we get
\( \sec^2 \theta = \left( x + \frac{1}{4x} \right)^2 \)
We know that
\( \tan^2 \theta = \sec^2 \theta - 1 \)
\( \Rightarrow \tan^2 \theta = \left( x + \frac{1}{4x} \right)^2 - 1 \)
\( \Rightarrow \tan^2 \theta = x^2 + \frac{1}{16x^2} + \frac{1}{2} - 1 \)
\( \Rightarrow \tan^2 \theta = x^2 + \frac{1}{16x^2} - \frac{1}{2} \)
\( \Rightarrow \tan^2 \theta = \left( x - \frac{1}{4x} \right)^2 \)
\( \Rightarrow \tan \theta = \pm \left( x - \frac{1}{4x} \right) \)
When \( \tan \theta = \left( x - \frac{1}{4x} \right) \)
Then, \( \sec \theta + \tan \theta = x + \frac{1}{4x} + x - \frac{1}{4x} = 2x \)
When \( \tan \theta = -\left( x - \frac{1}{4x} \right) \)
Then \( \sec \theta + \tan \theta = \left( x + \frac{1}{4x} \right) - \left( x - \frac{1}{4x} \right) = \frac{2}{4x} = \frac{1}{2x} \)
\( \sec \theta + \tan \theta = 2x \) or \( \frac{1}{2x} \). Hence Proved.

Assertion and Reasoning Based Questions

DIRECTIONS : In the following questions, a statement 1 is followed by statement 2. Mark the correct choice as :
(a) If both statement 1 and statement 2 are true and statement 2 is the correct explanation of statement 1.
(b) If both statement 1 and statement 2 are true, but statement 2 is not the correct explanation of statement 1.
(c) If statement 1 is true, but statement 2 is false.
(d) If statement 1 is false, but statement 2 is true.

Question. Statement 1 : In a right-angle \(\Delta ABC\), right-angled at B, AB = 4 and BC = 3. Then, value of \( \sec A = \frac{4}{5} \).
Statement 2 : ‘Secant of angle \(\theta\)’ is reverse of ‘cosecant of angle \(\theta\)’.
(a) A
(b) B
(c) C
(d) D
Answer: (b) Statement 1 is correct as by Pythagoras theorem, hypotenuse will be 5. So, the value of sec A will be \( \frac{4}{5} \) and statement 2 is also correct but not the correct reason for statement 1.

Question. Statement 1 : In a right angled \(\Delta ABC\), if \( \text{cosec } A = \sqrt{10} \) then, the value of cot A for the same triangle is \( \frac{2}{1} \).
Statement 2 : By applying Pythagoras theorem we can find the hypotenuse, if base and perpendicular of a right angled triangle is given. And then all trigonometric ratios can be calculated.
(a) A
(b) B
(c) C
(d) D
Answer: (d) Statement 1 is incorrect as value of cot A is \( \frac{3}{1} \) and statement 2 is a complete reason for calculating trigonometric ratio and it is correct.

Question. Statement 1 : The value of \( \tan 90^\circ \) is not-defined.
Statement 2 : ‘Tangent of angle \(\theta\)’ is the ratio of the ‘sine of angle \(\theta\)’ and ‘cosine of angle \(\theta\)’.
(a) A
(b) B
(c) C
(d) D
Answer: (a) \( \tan 90^\circ = \infty \) as \( \sin 90^\circ = 1 \) and \( \cos 90^\circ = 0 \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
\(\therefore\) statement 2 gives correct reason for statement 1.

Question. Statement 1 : If A and B are complementary angles then \( \sin A = \cos B \) and vice-versa.
Statement 2 : Two angles are said to be complementary if their sum is 180 i.e., \(\theta\) and \( (180^\circ - \theta) \) are complementary angles for an angle \(\theta\).
(a) A
(b) B
(c) C
(d) D
Answer: (c) For statement 1 : \( A + B = 90^\circ \)
\(\therefore A = 90^\circ - B \)
\( \Rightarrow \sin A = \sin (90^\circ - B) \)
\( \Rightarrow \sin A = \cos B \)
But statement 2 is not correct as sum of complementary angles is \( 90^\circ \) not \( 180^\circ \).