CBSE Class 10 Maths HOTs Introduction to Trigonometry Set 06

Very Short Answer Type Questions

Question. Write the value of \( \sin^2 30^\circ + \cos^2 60^\circ \). 
Answer: \( = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 \)
\( = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \)

Question. If \( \sin x + \cos y = 1; x = 30^\circ \) and \( y \) is an acute angle, find the value of \( y \). 
Answer: Given, \( \sin x + \cos y = 1 \) and \( x = 30^\circ \)
\( \therefore \sin 30^\circ + \cos y = 1 \)
\( \implies \cos y = 1 - \frac{1}{2} = \frac{1}{2} \)
\( \implies \cos y = \cos 60^\circ \)
\( \therefore y = 60^\circ \)
Hence, the value of \( y \) is 60º

Question. \( \sin^2 60^\circ + 2 \tan 45^\circ - \cos^2 30^\circ \) 
Answer: \( \sin^2 60^\circ + 2 \tan 45^\circ - \cos^2 30^\circ \)
Since, \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
\( \tan 45^\circ = 1 \)
\( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
\( = \left(\frac{\sqrt{3}}{2}\right)^2 + 2 \times 1 - \left(\frac{\sqrt{3}}{2}\right)^2 \)
\( = \frac{3}{4} + 2 - \frac{3}{4} = 2 \)
Hence, the value is 2.

Question. If \( \sin A = \frac{3}{4} \), calculate \( \sec A \). 
Answer: Given: \( \sin A = \frac{3}{4} \)
Since, \( \sin^2 A + \cos^2 A = 1 \)
\( \cos^2 A = 1 - \sin^2 A \)
or \( \cos A = \sqrt{1 - \sin^2 A} \)
\( = \sqrt{1 - \left(\frac{3}{4}\right)^2} \)
\( = \sqrt{1 - \frac{9}{16}} \)
\( = \frac{\sqrt{7}}{4} \)
\( \therefore \sec A = \frac{1}{\cos A} = \frac{4}{\sqrt{7}} \)
Hence, the value of \( \sec A \) is \( \frac{4}{\sqrt{7}} \).

Question. If \( (1 + \cos A) (1 - \cos A) = \frac{3}{4} \), find value of \( \sec A \). 
Answer: \( (1 + \cos A) (1 - \cos A) = \frac{3}{4} \)
\( \implies 1 - \cos^2 A = \frac{3}{4} \)
\( \implies \cos^2 A = 1 - \frac{3}{4} \)
\( \implies \sec^2 A = 4 \)
\( \implies \sec A = \pm 2 \)

Question. If \( \tan A = 1 (0^\circ < A < 90^\circ) \) and \( \cos B = \frac{1}{\sqrt{2}} (0^\circ < B < 90^\circ) \), then find \( \cos (A + B) \)
Answer: \( \tan A = 1 \implies A = 45^\circ \)
and \( \cos B = \frac{1}{\sqrt{2}} \implies B = 45^\circ \)
So, \( \cos (A + B) = \cos (45^\circ + 45^\circ) \)
\( = \cos 90^\circ = 0 \)

Question. Evaluate : \( 2 \sec 30^\circ \times \tan 60^\circ \). 
Answer: \( 2 \sec 30^\circ \times \tan 60^\circ \)
\( = 2 \times \frac{2}{\sqrt{3}} \times \sqrt{3} \)
\( = 4 \)

Short Answer (SA-I) Type Questions

Question. Prove that \( 1 + \frac{\cot^2 \alpha}{1 + \csc \alpha} = \csc \alpha \) 
Answer: L.H.S. \( = 1 + \frac{\cot^2 \alpha}{1 + \csc \alpha} \)
\( = 1 + \frac{\csc^2 \alpha - 1}{1 + \csc \alpha} \)
\( = 1 + \frac{(\csc \alpha - 1)(\csc \alpha + 1)}{1 + \csc \alpha} \)
\( = 1 + (\csc \alpha - 1) \)
\( = \csc \alpha = \text{RHS} \)

Question. Show that \( \tan^4 \theta + \tan^2 \theta = \sec^4 \theta - \sec^2 \theta \) 
Answer: We know that
\( \sec^4 \theta - \tan^4 \theta \)
\( = (\sec^2 \theta + \tan^2 \theta) (\sec^2 \theta - \tan^2 \theta) \)
\( = (\sec^2 \theta + \tan^2 \theta) [1 + \tan^2 \theta - \tan^2 \theta] \)
\( = \sec^2 \theta + \tan^2 \theta \)
\( \implies \sec^4 \theta - \sec^2 \theta = \tan^4 \theta + \tan^2 \theta \)

Question. Find \( A \) and \( B \), if \( \sin (A + 2B) = \frac{\sqrt{3}}{2} \) and \( \cos (A + B) = \frac{1}{2} \) 
Answer: Given : \( \sin (A + 2B) = \sin 60^\circ \)
[\( \because \sin 60^\circ = \frac{\sqrt{3}}{2} \)]
\( \therefore A + 2B = 60^\circ \text{ ...(i)} \)
\( \cos (A + B) = \cos 60^\circ \)
[\( \because \cos 60^\circ = \frac{1}{2} \)]
\( \therefore A + B = 60^\circ \text{ ...(ii)} \)
Subtracting equation (i) and (ii)
\( B = 0^\circ \)
Putting the value of \( B \) in equation (ii), we get,
\( A = 60^\circ - 0^\circ = 60^\circ \)
So, \( A = 60^\circ \text{ and } B = 0^\circ \).

Question. If \( x \cos \theta - y \sin \theta = a, x \sin \theta + y \cos \theta = b \), prove that \( x^2 + y^2 = a^2 + b^2 \).
Answer: \( x \cos \theta - y \sin \theta = a \text{ ...(i)} \)
\( x \sin \theta + y \cos \theta = b \text{ ...(ii)} \)
Squaring and adding both equations (i) and (ii), we get
\( (x \cos \theta - y \sin \theta)^2 + (x \sin \theta + y \cos \theta)^2 = a^2 + b^2 \)
\( \implies x^2 \cos^2 \theta + y^2 \sin^2 \theta - 2xy \sin \theta \cos \theta + x^2 \sin^2 \theta + y^2 \cos^2 \theta + 2xy \sin \theta \cos \theta = a^2 + b^2 \)
\( \implies x^2 (\cos^2 \theta + \sin^2 \theta) + y^2 (\sin^2 \theta + \cos^2 \theta) = a^2 + b^2 \)
\( \implies x^2 + y^2 = a^2 + b^2 \)

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: \( x = a \cos^3 \theta, y = b \sin^3 \theta \)
LHS \( = \left(\frac{x}{a}\right)^{2/3} + \left(\frac{y}{b}\right)^{2/3} \)
\( = \left(\frac{a \cos^3 \theta}{a}\right)^{2/3} + \left(\frac{b \sin^3 \theta}{b}\right)^{2/3} \)
\( = (\cos \theta)^{3 \times 2/3} + (\sin \theta)^{3 \times 2/3} \)
\( = \cos^2 \theta + \sin^2 \theta \)
\( = 1 \) [\( \because \sin^2 \theta + \cos^2 \theta = 1 \)]
LHS = RHS. Hence, proved.

Question. The shadow of a 5 m long stick is 2 m long. At the same time, find the length of the shadow of a 12.5 m high tree is
Answer: Let the length of a shadow of 12.5 m high tree be \( x \) m.
Now, ratio of lengths of objects = Ratio of lengths of their shadows
\( \frac{5}{12.5} = \frac{2}{x} \)
\( x = \frac{2 \times 12.5}{5} = \frac{25}{5} = 5 \text{ m} \)

Question. Evaluate: \( (\sin^4 60^\circ + \sec^4 30^\circ) - 2 (\cos^2 45^\circ - \sin^2 90^\circ) \).
Answer: Expression \( = (\sin^4 60^\circ + \sec^4 30^\circ) - 2 (\cos^2 45^\circ - \sin^2 90^\circ) \)
\( = \left[ \left(\frac{\sqrt{3}}{2}\right)^4 + \left(\frac{2}{\sqrt{3}}\right)^4 \right] - 2 \left[ \left(\frac{1}{\sqrt{2}}\right)^2 - (1)^2 \right] \)
\( = \left( \frac{9}{16} + \frac{16}{9} \right) - 2 \left( \frac{1}{2} - 1 \right) \)
\( = \frac{337}{144} + 1 \)
\( = 2 \frac{49}{144} + 1 \)
\( = 3 \frac{49}{144} \)

Question. If \( a \cos \theta - b \sin \theta = c \), prove that \( a \sin \theta + b \cos \theta = \pm \sqrt{a^2 + b^2 - c^2} \).
Answer: \( a \cos \theta - b \sin \theta = c \)
On squaring both sides, we get
\( (a \cos \theta - b \sin \theta)^2 = c^2 \)
\( \implies a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \cos \theta \sin \theta = c^2 \)
\( \implies a^2(1 - \sin^2 \theta) + b^2(1 - \cos^2 \theta) - 2ab \cos \theta \sin \theta = c^2 \)
\( \implies a^2 - a^2 \sin^2 \theta + b^2 - b^2 \cos^2 \theta - 2ab \cos \theta \sin \theta = c^2 \)
\( \implies a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \cos \theta \sin \theta = a^2 + b^2 - c^2 \)
\( \implies (a \sin \theta + b \cos \theta)^2 = a^2 + b^2 - c^2 \)
\( \implies a \sin \theta + b \cos \theta = \pm \sqrt{a^2 + b^2 - c^2} \)

Question. Simplify \( (1 + \tan^2 \theta) (1 - \sin \theta) (1 + \sin \theta) \). 
Answer: We know that \( 1 + \tan^2 \theta = \sec^2 \theta \)
\( (1 + \tan^2 \theta) (1 - \sin \theta) (1 + \sin \theta) \)
\( = (1 + \tan^2 \theta) (1 - \sin^2 \theta) \) [\( \because (a - b) (a + b) = a^2 - b^2 \)]
\( = \sec^2 \theta \cdot \cos^2 \theta \) [\( \because 1 + \tan^2 \theta = \sec^2 \theta \text{ and } \cos^2 \theta = 1 - \sin^2 \theta \)]
\( = \frac{1}{\cos^2 \theta} \cdot \cos^2 \theta = 1 \) [\( \because \sec \theta = \frac{1}{\cos \theta} \)]

Short Answer (SA-II) Type Questions

Question. Prove that : \( \frac{\sin \theta - \cos \theta + 1}{\cos \theta + \sin \theta - 1} = \frac{1}{\sec \theta - \tan \theta} \)
Answer: LHS \( = \frac{\sin \theta - \cos \theta + 1}{\cos \theta + \sin \theta - 1} \)
\( = \frac{\tan \theta - 1 + \sec \theta}{1 + \tan \theta - \sec \theta} \) [Dividing \( N_r \) and \( D_r \) by \( \cos \theta \)]
\( = \frac{(\tan \theta + \sec \theta) - (\sec^2 \theta - \tan^2 \theta)}{1 + \tan \theta - \sec \theta} \)
[\( \because 1 + \tan^2 \theta = \sec^2 \theta \)]
\( = \frac{(\tan \theta + \sec \theta) [1 - \sec \theta + \tan \theta]}{1 + \tan \theta - \sec \theta} \)
\( = \tan \theta + \sec \theta \)
\( = (\tan \theta + \sec \theta) \times \frac{\sec \theta - \tan \theta}{\sec \theta - \tan \theta} \)
\( = \frac{\sec^2 \theta - \tan^2 \theta}{\sec \theta - \tan \theta} \)
\( = \frac{1}{\sec \theta - \tan \theta} = \text{RHS} \)

Question. Prove that : \( \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sec A + \tan A \) 
Answer: LHS \( = \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} \)
\( = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} = \frac{1 + \sin A}{\sqrt{\cos^2 A}} \)
\( = \frac{1 + \sin A}{\cos A} = \frac{1}{\cos A} + \frac{\sin A}{\cos A} \)
\( = \sec A + \tan A \).

Question. Prove that \( \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \csc \theta \) 
Answer: We will use identities \( \sin^2 \theta + \cos^2 \theta = 1 \) and \( \frac{1}{\sin \theta} = \csc \theta \)
L.H.S. \( = \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} \)
\( = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{(1 + \cos \theta) \sin \theta} = \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2 \cos \theta}{\sin \theta (1 + \cos \theta)} \)
\( = \frac{(\sin^2 \theta + \cos^2 \theta) + 1 + 2 \cos \theta}{\sin \theta (1 + \cos \theta)} \)
\( = \frac{1 + 1 + 2 \cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} \)
\( = \frac{2}{\sin \theta} = 2 \csc \theta \)
\( = \text{R.H.S.} \) Hence, proved

Question. If \( \sin \theta + \cos \theta = \sqrt{2} \), prove that \( \tan \theta + \cot \theta = 2 \). 
Answer: Given; \( \sin \theta + \cos \theta = \sqrt{2} \),
On squaring both sides, we get:
\( \implies (\sin \theta + \cos \theta)^2 = (\sqrt{2})^2 \)
\( \implies \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 2 \)
\( \implies 1 + 2 \sin \theta \cos \theta = 2 \)
or \( \sin 2\theta = 1 \text{ ...(ii)} \)
\( \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \)
\( = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{2}{2 \sin \theta \cos \theta} = \frac{2}{\sin 2\theta} \)
\( = \frac{2}{1} = 2 \) (from (ii))

Question. Prove that : \( \frac{\cot \theta + \csc \theta - 1}{\cot \theta - \csc \theta + 1} = \frac{1 + \cos \theta}{\sin \theta} \) 
Answer: LHS \( = \frac{\cot \theta + \csc \theta - 1}{\cot \theta - \csc \theta + 1} \)
\( = \frac{(\cot \theta + \csc \theta) - (\csc^2 \theta - \cot^2 \theta)}{\cot \theta - \csc \theta + 1} \) [\( \because 1 + \cot^2 \theta = \csc^2 \theta \)]
\( = \frac{(\cot \theta + \csc \theta) (1 - \csc \theta + \cot \theta)}{\cot \theta - \csc \theta + 1} \)
\( = \cot \theta + \csc \theta \)
\( = \frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta} = \frac{1 + \cos \theta}{\sin \theta} = \text{RHS} \)
Hence, proved

Question. Prove that : \( 2 (\sin^6 \theta + \cos^6 \theta) - 3 (\sin^4 \theta + \cos^4 \theta) + 1 = 0 \). 
Answer: We know that:
\( \sin^2 \theta + \cos^2 \theta = 1 \)
So, \( (\sin^2 \theta + \cos^2 \theta)^2 = 1^2 \)
i.e., \( \sin^4 \theta + \cos^4 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta \text{ ...(i)} \)
Also, \( (\sin^2 \theta + \cos^2 \theta)^3 = 1^3 \)
i.e., \( \sin^6 \theta + \cos^6 \theta = 1 - 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) \)
\( = 1 - 3 \sin^2 \theta \cos^2 \theta \text{ ...(ii)} \)
Using (i) and (ii), we get
\( 2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) + 1 \)
\( = 2[1 - 3 \sin^2 \theta \cos^2 \theta] - 3(1 - 2 \sin^2 \theta \cos^2 \theta) + 1 \)
\( = 2 - 3 + 1 = 0 \)

Question. If \( \sin \theta + \cos \theta = \sqrt{3} \), then prove that \( \tan \theta + \cot \theta = 1 \). 
Answer: It is given that \( \sin \theta + \cos \theta = \sqrt{3} \)
\( \implies (\sin \theta + \cos \theta)^2 = 3 \)
\( \implies 1 + 2 \sin \theta \cos \theta = 3 \text{ ..(i)} \)
Hence, \( \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} \)
\( = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} \)
\( = \frac{1}{1} \) (By (i)) \( = 1 \)

Question. Prove that : \( (\sin^4 \theta - \cos^4 \theta + 1) \csc^2 \theta = 2 \). 
Answer: LHS \( = (\sin^4 \theta - \cos^4 \theta + 1) \csc^2 \theta \)
\( = [(\sin^2 \theta + \cos^2 \theta) (\sin^2 \theta - \cos^2 \theta) + 1] \csc^2 \theta \)
\( = [(\sin^2 \theta - \cos^2 \theta) + 1] \csc^2 \theta \)
\( = [\sin^2 \theta + (1 - \cos^2 \theta)] \csc^2 \theta \)
\( = [2 \sin^2 \theta] \csc^2 \theta \)
\( = 2 (\sin^2 \theta \csc^2 \theta) = 2 \times 1 = 2 \text{ (RHS)} \)

Question. Prove that : \( \frac{2 \cos^3 \theta - \cos \theta}{\sin \theta - 2 \sin^3 \theta} = \cot \theta \) 
Answer: LHS \( = \frac{2 \cos^3 \theta - \cos \theta}{\sin \theta - 2 \sin^3 \theta} \)
\( = \frac{\cos \theta [2 \cos^2 \theta - 1]}{\sin \theta [1 - 2 \sin^2 \theta]} \)
\( = \cot \theta \cdot \left[ \frac{\cos^2 \theta + \cos^2 \theta - 1}{1 - \sin^2 \theta - \sin^2 \theta} \right] \)
\( = \cot \theta \cdot \left[ \frac{\cos^2 \theta - \sin^2 \theta}{\cos^2 \theta - \sin^2 \theta} \right] \)
\( = \cot \theta = \text{RHS} \)

Question. \( \frac{\tan \theta}{1 - \tan \theta} - \frac{\cot \theta}{1 - \cot \theta} = \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \) 
Answer: To prove : \( \frac{\tan \theta}{1 - \tan \theta} - \frac{\cot \theta}{1 - \cot \theta} = \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta} \)
Proof : L.H.S. \( = \frac{\tan \theta}{1 - \tan \theta} - \frac{\cot \theta}{1 - \cot \theta} \)
\( = \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\sin \theta}{\cos \theta}} - \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\cos \theta}{\sin \theta}} \)
\( = \frac{\sin \theta}{\cos \theta - \sin \theta} - \frac{\cos \theta}{\sin \theta - \cos \theta} \)
\( = \frac{\sin \theta}{\cos \theta - \sin \theta} + \frac{\cos \theta}{\cos \theta - \sin \theta} \)
\( = \frac{\sin \theta + \cos \theta}{\cos \theta - \sin \theta} \) = RHS Hence proved

Question. If \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \), show that \( \cos \theta - \sin \theta = \sqrt{2} \sin \theta \) 
Answer: Given : \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \)
To prove : \( \cos \theta - \sin \theta = \sqrt{2} \sin \theta \)
Proof : \( \cos \theta + \sin \theta = \sqrt{2} \cos \theta \)
\( \implies \sin \theta = (\sqrt{2} - 1) \cos \theta \)
Multiply both sides by \( (\sqrt{2} + 1) \), we get:
\( \implies (\sqrt{2} + 1) \sin \theta = (\sqrt{2} + 1)(\sqrt{2} - 1) \cos \theta \)
\( \implies \sqrt{2} \sin \theta + \sin \theta = (2 - 1) \cos \theta \)
\( \implies \sqrt{2} \sin \theta + \sin \theta = \cos \theta \)
\( \implies \cos \theta - \sin \theta = \sqrt{2} \sin \theta \) Hence, proved

Question. Prove that: \( (\sin \theta + 1 + \cos \theta) (\sin \theta - 1 + \cos \theta) \cdot \sec \theta \csc \theta = 2 \)
Answer: Proof : LHS
\( = (\sin \theta + \cos \theta + 1) (\sin \theta + \cos \theta - 1) \cdot \sec \theta \csc \theta \)
\( = [(\sin \theta + \cos \theta)^2 - (1)^2] \cdot \sec \theta \csc \theta \)
\( = [\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta - 1] \sec \theta \csc \theta \)
\( = [1 + 2 \sin \theta \cos \theta - 1] \sec \theta \csc \theta \)
\( = (2 \sin \theta \cos \theta) \cdot \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta} \)
\( = 2 = \text{RHS} \). Hence proved

Question. Prove that: \( \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} + \sqrt{\frac{\sec \theta + 1}{\sec \theta - 1}} = 2 \csc \theta \)
Answer: Proof : LHS : \( \sqrt{\frac{\sec \theta - 1}{\sec \theta + 1}} + \sqrt{\frac{\sec \theta + 1}{\sec \theta - 1}} \)
\( = \frac{(\sqrt{\sec \theta - 1})^2 + (\sqrt{\sec \theta + 1})^2}{\sqrt{(\sec \theta + 1)(\sec \theta - 1)}} \)
\( = \frac{\sec \theta - 1 + \sec \theta + 1}{\sqrt{\sec^2 \theta - 1}} \)
\( = \frac{2 \sec \theta}{\sqrt{\tan^2 \theta}} = \frac{2 \sec \theta}{\tan \theta} \)
\( = 2 \times \frac{1}{\cos \theta} \times \frac{\cos \theta}{\sin \theta} \)
\( = \frac{2}{\sin \theta} = 2 \csc \theta = \text{RHS} \) Hence proved

Question. Prove that \( (\sin \theta + \csc \theta)^2 + (\cos \theta + \sec \theta)^2 = 7 + \tan^2 \theta + \cot^2 \theta \). 
Answer: LHS \( = (\sin \theta + \csc \theta)^2 + (\cos \theta + \sec \theta)^2 \)
\( = \sin^2 \theta + \csc^2 \theta + 2 \sin \theta \csc \theta + \cos^2 \theta + \sec^2 \theta + 2 \cos \theta \sec \theta \)
\( = (\sin^2 \theta + \cos^2 \theta) + (\csc^2 \theta + \sec^2 \theta) + 2 + 2 \)
\( = 1 + (1 + \cot^2 \theta) + (1 + \tan^2 \theta) + 4 \)
\( = 7 + \tan^2 \theta + \cot^2 \theta = \text{RHS} \) Hence, Proved.

Question. Prove that \( (1 + \cot A - \csc A) (1 + \tan A + \sec A) = 2 \) 
Answer: LHS \( = (1 + \cot A - \csc A) (1 + \tan A + \sec A) \)
\( = \left( 1 + \frac{\cos A}{\sin A} - \frac{1}{\sin A} \right) \left( 1 + \frac{\sin A}{\cos A} + \frac{1}{\cos A} \right) \)
\( = \left( \frac{\sin A + \cos A - 1}{\sin A} \right) \left( \frac{\cos A + \sin A + 1}{\cos A} \right) \)
\( = \frac{(\sin A + \cos A)^2 - (1)^2}{\sin A \cos A} \)
\( = \frac{\sin^2 A + \cos^2 A + 2 \sin A \cos A - 1}{\sin A \cos A} \)
\( = \frac{1 + 2 \sin A \cos A - 1}{\sin A \cos A} = 2 = \text{RHS} \) Hence, proved.

Question. If \( 4 \tan \theta = 3 \), evaluate \( \frac{4 \sin \theta - \cos \theta + 1}{4 \sin \theta + \cos \theta - 1} \) 
Answer: Given, \( 4 \tan \theta = 3 \)
\( \implies \tan \theta = \frac{3}{4} \text{ and } \sec^2 \theta = 1 + \tan^2 \theta \)
\( \therefore \sec \theta = \sqrt{1 + \frac{9}{16}} = \frac{5}{4} \)
Dividing the numerator and denominator by \( \cos \theta \):
\( = \frac{4 \tan \theta - 1 + \sec \theta}{4 \tan \theta + 1 - \sec \theta} \)
\( = \frac{4 \times \frac{3}{4} - 1 + \frac{5}{4}}{4 \times \frac{3}{4} + 1 - \frac{5}{4}} = \frac{3 - 1 + \frac{5}{4}}{3 + 1 - \frac{5}{4}} = \frac{2 + \frac{5}{4}}{4 - \frac{5}{4}} = \frac{13}{11} \)
Hence, the required value is \( \frac{13}{11} \).

Question. Using the formula \( \cos 2\theta = 2 \cos^2 \theta - 1 \), find the value of \( \cos 30^\circ \), it is being given that \( \cos 60^\circ = \frac{1}{2} \). 
Answer: Given, \( \cos 2\theta = 2 \cos^2 \theta - 1 \)
Let \( \theta = 30^\circ \)
Then, \( \cos(2 \times 30^\circ) = 2 \cos^2 30^\circ - 1 \)
\( \implies \cos 60^\circ = 2 \cos^2 30^\circ - 1 \)
\( \implies \frac{1}{2} + 1 = 2 \cos^2 30^\circ \)
\( \implies 2 \cos^2 30^\circ = \frac{3}{2} \)
\( \implies \cos^2 30^\circ = \frac{3}{4} \text{ and } \cos 30^\circ = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \)

Question. If \( \sin \theta + \cos \theta = \sqrt{3} \), then prove that \( \tan \theta + \cot \theta = 1 \).
Answer: \( \sin \theta + \cos \theta = \sqrt{3} \)
\( \implies (\sin \theta + \cos \theta)^2 = 3 \)
\( \implies 1 + 2 \sin \theta \cos \theta = 3 \)
\( \implies \sin \theta \cos \theta = 1 \)
\( \therefore \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{1} = 1 \)

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Question. Prove the following identity: \( \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A \)
Answer: LHS = \( \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} \)
\( = \frac{\cos^2 A + (1 + \sin A)^2}{\cos A(1 + \sin A)} \)
\( = \frac{\cos^2 A + 1 + \sin^2 A + 2 \sin A}{\cos A(1 + \sin A)} \)
\( = \frac{2 + 2 \sin A}{\cos A(1 + \sin A)} \) {Since \( \cos^2 A + \sin^2 A = 1 \)}
\( = \frac{2(1 + \sin A)}{\cos A(1 + \sin A)} \)
\( = \frac{2}{\cos A} \)
\( = 2 \sec A = \text{RHS.} \)

Question. Prove that: \( \sec^2 \theta + \text{cosec}^2 \theta = \sec^2 \theta \cdot \text{cosec}^2 \theta \)
Answer: LHS = \( \sec^2 \theta + \text{cosec}^2 \theta \)
\( = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} \)
\( = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta \sin^2 \theta} \)
\( = \frac{1}{\cos^2 \theta \sin^2 \theta} \)
\( = \sec^2 \theta \cdot \text{cosec}^2 \theta = \text{RHS.} \)

Question. If \( 2 \sin^2 \theta - \cos^2 \theta = 2 \), find the value of \( \theta \).
Answer: Given: \( 2 \sin^2 \theta - \cos^2 \theta = 2 \)
\( \implies 2 \sin^2 \theta - (1 - \sin^2 \theta) = 2 \) {Since \( \cos^2 \theta = 1 - \sin^2 \theta \)}
\( \implies 2 \sin^2 \theta - 1 + \sin^2 \theta = 2 \)
\( \implies 3 \sin^2 \theta = 3 \)
\( \implies \sin^2 \theta = 1 \)
\( \implies \sin \theta = 1 \)
\( \implies \sin \theta = \sin 90^\circ \) {Since \( \sin 90^\circ = 1 \)}
\( \implies \theta = 90^\circ \)

Long Answer Type Question

Question. If \( 1 + \sin^2 \theta = 3 \sin \theta \cos \theta \), then prove that \( \tan \theta = 1 \) or \( \frac{1}{2} \).
Answer: To solve the equation in \( \theta \), we have to convert into 1 trigonometric ratio.
Given: \( 1 + \sin^2 \theta = 3 \sin \theta \cos \theta \)
To prove: \( \tan \theta = 1 \) or \( \frac{1}{2} \)
Proof: \( 1 + \sin^2 \theta = 3 \sin \theta \cos \theta \)
Dividing both sides by \( \sin^2 \theta \)
\( \implies \frac{1 + \sin^2 \theta}{\sin^2 \theta} = \frac{3 \sin \theta \cos \theta}{\sin^2 \theta} \)
\( \implies \frac{1}{\sin^2 \theta} + 1 = \frac{3 \cos \theta}{\sin \theta} \)
\( [\because \frac{1}{\sin \theta} = \csc \theta \text{ and } \cot \theta = \frac{\cos \theta}{\sin \theta}] \)
\( \implies \csc^2 \theta + 1 = 3 \cot \theta \)
\( \implies \cot^2 \theta + 1 + 1 = 3 \cot \theta \)
\( [\because \csc^2 \theta = 1 + \cot^2 \theta] \)
\( \implies \cot^2 \theta - 3 \cot \theta + 2 = 0 \)
\( \implies \cot^2 \theta - 2 \cot \theta - \cot \theta + 2 = 0 \)
\( \implies \cot \theta(\cot \theta - 2) - 1(\cot \theta - 2) = 0 \)
\( \implies (\cot \theta - 2)(\cot \theta - 1) = 0 \)
\( \implies \cot \theta = 1 \text{ or } 2 \)
\( \implies \tan \theta = 1 \text{ or } \frac{1}{2} \) Hence, proved.

Question. A boy standing on a horizontal plane finds a bird flying at a distance of 100 m from him at an elevation of \( 30^\circ \). A girl standing on the roof of a 20 m high building, find the elevation of the same bird to be \( 45^\circ \). The boy and the girl are on the opposite sides of the bird. Find the distance of the bird the girl. (Given \( \sqrt{2} = 1.414 \))
Answer: Let \( C \) be the position of the bird, \( A \) be the position of the boy and \( ED \) be the building at which roof girl is standing and '\( E \)' be the position of girl.
Let \( BC \perp AB \) and \( EF \perp BC \).
Then, \( AC = 100 \) m, \( \angle CAB = 30^\circ \), \( ED = 20 \) m, \( \angle CEF = 45^\circ \).
From right \( \triangle CBA \);
\( \sin 30^\circ = \frac{BC}{AC} = \frac{BC}{100} \)
\( \implies \frac{1}{2} = \frac{BC}{100} \)
\( \implies BC = 50 \) m
Now \( CF = BC - BF = BC - ED \)
\( = 50 - 20 = 30 \) m
From right \( \triangle CFE \),
\( \sin 45^\circ = \frac{CF}{CE} \)
\( \implies \frac{1}{\sqrt{2}} = \frac{30}{CE} \)
\( \implies CE = 30\sqrt{2} \) m
\( \implies CE = 30 \times 1.414 = 42.42 \) m
Hence, the distance of the bird from the girl = 42.42 m.

Question. If \( \sin \theta + \cos \theta = p \) and \( \sec \theta + \csc \theta = q \), then prove that \( q(p^2 - 1) = 2p \).
Answer: Given that:
\( \sin \theta + \cos \theta = p \)
\( \sec \theta + \csc \theta = q \)
To prove: \( q(p^2 - 1) = 2p \)
Proof: \( \sin \theta + \cos \theta = p \) ...(i)
\( \sec \theta + \csc \theta = q \)
\( \implies \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = q \)
\( [\because \sec \theta = \frac{1}{\cos \theta} \text{ and } \csc \theta = \frac{1}{\sin \theta}] \)
\( \implies \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = q \)
\( \implies \frac{p}{\sin \theta \cos \theta} = q \) [Using eqn (i)]
\( \implies \sin \theta \cos \theta = \frac{p}{q} \) ...(ii)
It is given that \( \sin \theta + \cos \theta = p \)
On squaring both sides, we get \( (\sin \theta + \cos \theta)^2 = p^2 \)
\( \implies \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = p^2 \)
\( \implies 1 + 2\sin \theta \cos \theta = p^2 \)
\( \implies 1 + 2\frac{p}{q} = p^2 \) [From (ii)]
\( \implies q + 2p = p^2 q \)
\( \implies p^2 q - q = 2p \)
\( \implies q(p^2 - 1) = 2p \) Hence, proved.

Question. Prove that : \( \frac{(1 + \cot \theta + \tan \theta)(\sin \theta - \cos \theta)}{\sec^3 \theta - \csc^3 \theta} = \sin^2 \theta \cos^2 \theta \).
Answer: Proof:
L.H.S. = \( \frac{(1 + \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta})(\sin \theta - \cos \theta)}{\frac{1}{\cos^3 \theta} - \frac{1}{\sin^3 \theta}} \)
\( = \frac{\frac{(\sin \theta \cos \theta + \cos^2 \theta + \sin^2 \theta)(\sin \theta - \cos \theta)}{\sin \theta \cos \theta}}{\frac{\sin^3 \theta - \cos^3 \theta}{\cos^3 \theta \sin^3 \theta}} \)
\( = \frac{(1 + \sin \theta \cos \theta)(\sin \theta - \cos \theta) \cos^2 \theta \sin^2 \theta}{\sin^3 \theta - \cos^3 \theta} \)
\( = \frac{(1 + \sin \theta \cos \theta)(\sin \theta - \cos \theta) \cos^2 \theta \sin^2 \theta}{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)} \)
\( [\because a^3 - b^3 = (a - b)(a^2 + ab + b^2)] \)
\( = \frac{(1 + \sin \theta \cos \theta)}{(1 + \sin \theta \cos \theta)} \times \cos^2 \theta \sin^2 \theta \)
\( = \sin^2 \theta \cos^2 \theta = RHS \). Hence proved.

Question. If \( \sec \theta + \tan \theta = m \), show that \( \frac{m^2 - 1}{m^2 + 1} = \sin \theta \).
Answer: Given : \( \sec \theta + \tan \theta = m \)
Proof : \( \sec \theta + \tan \theta = m \) (given) ...(i)
We know that \( \sec^2 \theta - \tan^2 \theta = 1 \)
\( \implies (\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1 \)
\( \implies (\sec \theta - \tan \theta) = \frac{1}{m} \) ...(ii)
From (i) and (ii), we get:
\( 2\sec \theta = m + \frac{1}{m} \) and \( 2\tan \theta = m - \frac{1}{m} \)
Now, \( \sin \theta = \frac{\tan \theta}{\sec \theta} = \frac{2\tan \theta}{2\sec \theta} = \frac{m - \frac{1}{m}}{m + \frac{1}{m}} \)
\( \implies \sin \theta = \frac{m^2 - 1}{m^2 + 1} \). Hence proved.

Question. Prove that : \( \frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \csc \theta \).
Answer: L.H.S. = \( \frac{\tan \theta}{1 - \frac{1}{\tan \theta}} + \frac{\frac{1}{\tan \theta}}{1 - \tan \theta} \)
\( = \frac{\tan^2 \theta}{\tan \theta - 1} + \frac{1}{\tan \theta(1 - \tan \theta)} = \frac{\tan^3 \theta - 1}{\tan \theta(\tan \theta - 1)} \)
\( = \frac{(\tan \theta - 1)(\tan^2 \theta + \tan \theta + 1)}{\tan \theta(\tan \theta - 1)} \)
\( = \frac{\tan^2 \theta + \tan \theta + 1}{\tan \theta} = \tan \theta + 1 + \cot \theta \)
\( = 1 + \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = 1 + \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} \)
\( = 1 + \frac{1}{\sin \theta \cos \theta} = 1 + \sec \theta \csc \theta = RHS \). Hence proved.

Question. Prove that : \( \frac{\sin \theta}{\cot \theta + \csc \theta} = 2 + \frac{\sin \theta}{\cot \theta - \csc \theta} \).
Answer: Consider \( \frac{\sin \theta}{\csc \theta + \cot \theta} - \frac{\sin \theta}{\csc \theta - \cot \theta} \)
\( = \sin \theta \left[ \frac{\csc \theta - \cot \theta - (\csc \theta + \cot \theta)}{\csc^2 \theta - \cot^2 \theta} \right] \)
\( = \sin \theta [ -2 \cot \theta ] / 1 = -2 \sin \theta \frac{\cos \theta}{\sin \theta} = -2 \cos \theta \). (Step mismatch in source image, re-evaluating)
Actually, \( \sin \theta (\csc \theta - \cot \theta + \csc \theta + \cot \theta) / 1 = \sin \theta (2 \csc \theta) = 2 \).
\( \therefore \frac{\sin \theta}{\csc \theta + \cot \theta} - \frac{\sin \theta}{\cot \theta - \csc \theta} = 2 \)
\( \implies \frac{\sin \theta}{\csc \theta + \cot \theta} = 2 + \frac{\sin \theta}{\cot \theta - \csc \theta} \). Hence Proved.

Question. Prove that \( \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} = \frac{1}{\sec A - \tan A} \).
Answer: LHS = \( \frac{\sin A - \cos A + 1}{\sin A + \cos A - 1} \). Dividing numerator and denominator by \( \cos A \):
\( = \frac{\tan A - 1 + \sec A}{\tan A + 1 - \sec A} = \frac{(\tan A + \sec A) - 1}{(\tan A - \sec A) + 1} \)
\( = \frac{(\tan A + \sec A) - (\sec^2 A - \tan^2 A)}{(\tan A - \sec A) + 1} \)
\( = \frac{(\tan A + \sec A) [1 - (\sec A - \tan A)]}{\tan A - \sec A + 1} = \sec A + \tan A \)
\( = \frac{(\sec A + \tan A)(\sec A - \tan A)}{\sec A - \tan A} = \frac{1}{\sec A - \tan A} = RHS \). Hence Proved.

Question. Prove that : \( \frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} = \tan A \).
Answer: LHS = \( \frac{\sin A(1 - 2\sin^2 A)}{\cos A(2\cos^2 A - 1)} \)
\( = \tan A \frac{1 - 2(1 - \cos^2 A)}{2\cos^2 A - 1} = \tan A \frac{2\cos^2 A - 1}{2\cos^2 A - 1} = \tan A = RHS \). Hence prove