CBSE Class 10 Mathematics Introduction to Trigonometry VBQs Set C

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
\( \sec^4 \theta - \tan^4 \theta = (\sec^2 \theta + \tan^2 \theta) (\sec^2 \theta - \tan^2 \theta) \)
\( = (\sec^2 \theta + \tan^2 \theta) [1 + \tan^2 \theta - \tan^2 \theta] \)
\( = \sec^2 \theta + \tan^2 \theta \)
\( \Rightarrow \sec^4 \theta - \sec^2 \theta = \tan^4 \theta + \tan^2 \theta \)

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

Question. A ladder is placed along a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is \( 2\text{ m} \) away from the wall and the ladder makes an angle of \( 60^\circ \) with the level of the ground. Find the height of the wall.
Answer: Let, \( AC \) be a ladder placed along a wall \( AB \). The foot of the ladder \( C \) is at the distance of \( 2\text{ m} \) from level ground and the ladder makes an angle of \( 60^\circ \) with the ground. Let, \( h \) be the height of the wall.
In \( \Delta ABC \),
\( \tan 60^\circ = \frac{AB}{BC} \)
\( \Rightarrow \sqrt{3} = \frac{h}{2} \)
\( \Rightarrow h = 2\sqrt{3}\text{ m} \)
Hence, the height of the wall is \( 2\sqrt{3}\text{ m} \).

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: \( x \cos \theta - y \sin \theta = a \) ...(i)
\( x \sin \theta + y \cos \theta = b \) ...(ii)
Squaring and adding both equations (i) and (ii), we get
\( (x \cos \theta - y \sin \theta)^2 + (x \sin \theta + y \cos \theta)^2 = a^2 + b^2 \)
\( \Rightarrow x^2 \cos^2 \theta + y^2 \sin^2 \theta - 2xy \sin \theta \cos \theta + x^2 \sin^2 \theta + y^2 \cos^2 \theta + 2xy \sin \theta \cos \theta = a^2 + b^2 \)
\( \Rightarrow x^2 (\cos^2 \theta + \sin^2 \theta) + y^2 (\sin^2 \theta + \cos^2 \theta) = a^2 + b^2 \)
\( \Rightarrow x^2 + y^2 = a^2 + b^2 \)

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

Question. The shadow of a \( 5\text{ m} \) long stick is \( 2\text{ m} \) long. At the same time, find the length of the shadow of a \( 12.5\text{ m} \) high tree.
Answer: Let the length of a shadow of \( 12.5\text{ m} \) high tree be \( x\text{ m} \).
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 \( = (\sin^4 60^\circ + \sec^4 30^\circ) - 2 (\cos^2 45^\circ - \sin^2 90^\circ) \)
\( = \left[ \left( \frac{\sqrt{3}}{2} \right)^4 + \left( \frac{2}{\sqrt{3}} \right)^4 \right] - 2 \left[ \left( \frac{1}{\sqrt{2}} \right)^2 - (1)^2 \right] \)
\( = \left( \frac{9}{16} + \frac{16}{9} \right) - 2 \left( \frac{1}{2} - 1 \right) \)
\( = \left( \frac{81 + 256}{144} \right) - 2 \left( -\frac{1}{2} \right) \)
\( = \frac{337}{144} + 1 \)
\( = \frac{337 + 144}{144} = \frac{481}{144} = 3 \frac{49}{144} \)

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
\( (a \cos \theta - b \sin \theta)^2 = c^2 \)
\( \Rightarrow a^2 \cos^2 \theta + b^2 \sin^2 \theta - 2ab \cos \theta \sin \theta = c^2 \)
\( \Rightarrow a^2(1 - \sin^2 \theta) + b^2(1 - \cos^2 \theta) - 2ab \cos \theta \sin \theta = c^2 \)
\( \Rightarrow a^2 - a^2 \sin^2 \theta + b^2 - b^2 \cos^2 \theta - 2ab \cos \theta \sin \theta = c^2 \)
\( \Rightarrow a^2 \sin^2 \theta + b^2 \cos^2 \theta + 2ab \cos \theta \sin \theta = a^2 + b^2 - c^2 \)
\( \Rightarrow (a \sin \theta + b \cos \theta)^2 = a^2 + b^2 - c^2 \)
\( \Rightarrow a \sin \theta + b \cos \theta = \pm \sqrt{a^2 + b^2 - c^2} \)

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

Question. The ratio of the height of a tower and the length of its shadow on the ground is \( \sqrt{3} : 1 \). What is the angle of elevation of the sun?
Answer: Let height of tower be \( AB \) and its shadow be \( BC \).
\( \frac{AB}{BC} = \tan \theta \)
But, \( \frac{AB}{BC} = \frac{\sqrt{3}}{1} \) [Given]
\( \therefore \tan \theta = \sqrt{3} \)
\( \tan \theta = \tan 60^\circ \)
\( \theta = 60^\circ \).

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: \( \text{LHS} = \frac{\sin \theta - \cos \theta + 1}{\cos \theta + \sin \theta - 1} \)
\( = \frac{\tan \theta - 1 + \sec \theta}{1 + \tan \theta - \sec \theta} \) [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: \( \text{LHS} = \sqrt{\frac{1 + \sin A}{1 - \sin A}} = \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} \)
\( = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} = \frac{1 + \sin A}{\sqrt{\cos^2 A}} \)
\( = \frac{1 + \sin A}{\cos A} \)
\( = \frac{1}{\cos A} + \frac{\sin A}{\cos A} \)
\( = \sec A + \tan A \)

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 \( \frac{1}{\sin \theta} = \csc \theta \).
\( \text{LHS} = \frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} \)
\( = \frac{\sin^2 \theta + (1 + \cos \theta)^2}{(1 + \cos \theta) \sin \theta} = \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2 \cos \theta}{\sin \theta (1 + \cos \theta)} \)
\( = \frac{(\sin^2 \theta + \cos^2 \theta) + 1 + 2 \cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{1 + 1 + 2 \cos \theta}{\sin \theta (1 + \cos \theta)} \)
\( = \frac{2 + 2 \cos \theta}{\sin \theta (1 + \cos \theta)} = \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} \)
\( = \frac{2}{\sin \theta} = 2 \csc \theta = \text{R.H.S.} \) 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:
\( (\sin \theta + \cos \theta)^2 = (\sqrt{2})^2 \)
\( \Rightarrow \sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta = 2 \)
\( \Rightarrow 1 + 2 \sin \theta \cos \theta = 2 \)
or \( \sin 2\theta = 1 \) ...(ii)
\( \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta} = \frac{2}{1} = 2 \) (from (ii))

Question. Prove that : \( \frac{\cot \theta + \csc \theta - 1}{\cot \theta - \csc \theta + 1} = \frac{1 + \cos \theta}{\sin \theta} \)
Answer: \( \text{LHS} = \frac{\cot \theta + \csc \theta - 1}{\cot \theta - \csc \theta + 1} \)
\( = \frac{(\cot \theta + \csc \theta) - (\csc^2 \theta - \cot^2 \theta)}{\cot \theta - \csc \theta + 1} \) \( [\because 1 + \cot^2 \theta = \csc^2 \theta] \)
\( = \frac{(\cot \theta + \csc \theta)(1 - \csc \theta + \cot \theta)}{\cot \theta - \csc \theta + 1} = \cot \theta + \csc \theta \)
\( = \frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta} = \frac{1 + \cos \theta}{\sin \theta} = \text{RHS} \) 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., \( \sin^4 \theta + \cos^4 \theta = 1 - 2 \sin^2 \theta \cos^2 \theta \) ...(ii)
Also, \( (\sin^2 \theta + \cos^2 \theta)^3 = 1^3 \)
i.e., \( \sin^6 \theta + \cos^6 \theta = 1 - 3 \sin^2 \theta \cos^2 \theta (\sin^2 \theta + \cos^2 \theta) = 1 - 3 \sin^2 \theta \cos^2 \theta \) ...(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 \( \sin \theta + \cos \theta = \sqrt{3} \)
\( \Rightarrow (\sin \theta + \cos \theta)^2 = 3 \)
\( \Rightarrow 1 + 2 \sin \theta \cos \theta = 3 \Rightarrow \sin \theta \cos \theta = 1 \) ..(i)
Hence,
\( \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{1}{1} = 1 \) (By (i))

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

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

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

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

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

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

Question. If \( 4 \tan \theta = 3 \), evaluate \( \frac{4 \sin \theta - \cos \theta + 1}{4 \sin \theta + \cos \theta - 1} \) 
Answer: Given, \( 4 \tan \theta = 3 \Rightarrow \tan \theta = \frac{3}{4} \) and \( 1 + \tan^2 \theta = \sec^2 \theta \).
\( \therefore \sec \theta = \sqrt{1 + \frac{9}{16}} = \frac{25}{16} = \frac{5}{4} \).
Dividing the numerator and denominator by \( \cos \theta \):
\( = \frac{4 \tan \theta - 1 + \sec \theta}{4 \tan \theta + 1 - \sec \theta} \). If we put the required values
\( = \frac{4 \times \frac{3}{4} - 1 + \frac{5}{4}}{4 \times \frac{3}{4} + 1 - \frac{5}{4}} = \frac{3 - 1 + \frac{5}{4}}{3 + 1 - \frac{5}{4}} = \frac{2 + \frac{5}{4}}{4 - \frac{5}{4}} = \frac{13}{11} \).
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, \( \cos(2 \times 30^\circ) = 2 \cos^2 30^\circ - 1 \)
\( \Rightarrow \cos 60^\circ = 2 \cos^2 30^\circ - 1 \)
\( \Rightarrow \frac{1}{2} + 1 = 2 \cos^2 30^\circ \Rightarrow 2 \cos^2 30^\circ = \frac{3}{2} \)
\( \Rightarrow \cos^2 30^\circ = \frac{3}{4} \) 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: \( \sin \theta + \cos \theta = \sqrt{3} \Rightarrow (\sin \theta + \cos \theta)^2 = 3 \Rightarrow 1 + 2 \sin \theta \cos \theta = 3 \Rightarrow 2 \sin \theta \cos \theta = 2 \Rightarrow \sin \theta \cos \theta = 1 \).
\( \therefore \tan \theta + \cot \theta = \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{1} = 1 \) [CBSE Marking Scheme 2019]

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

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

Question. If \( 2 \sin^2 \theta - \cos^2 \theta = 2 \), find the value of \( \theta \). 
Answer: Given: \( 2 \sin^2 \theta - \cos^2 \theta = 2 \Rightarrow 2 \sin^2 \theta - (1 - \sin^2 \theta) = 2 \)
\( [\because \cos^2 \theta = 1 - \sin^2 \theta] \)
\( \Rightarrow 2 \sin^2 \theta - 1 + \sin^2 \theta = 2 \Rightarrow 3 \sin^2 \theta = 3 \Rightarrow \sin^2 \theta = 1 \)
\( \Rightarrow \sin \theta = 1 \Rightarrow \sin \theta = \sin 90^\circ \Rightarrow \theta = 90^\circ \)

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)
\(\sec \theta + \csc \theta = q\)
\(\Rightarrow \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = q\)
\([ \because \sec \theta = \frac{1}{\cos \theta} \text{ and } \csc \theta = \frac{1}{\sin \theta} ]\)
\(\Rightarrow \frac{\sin \theta + \cos \theta}{\sin \theta \cdot \cos \theta} = q\)
\(\Rightarrow \frac{p}{\sin \theta \cdot \cos \theta} = q\) [Using eqn (i)]
\(\Rightarrow \sin \theta \cdot \cos \theta = \frac{p}{q}\) ...(ii)
It is given that \(\sin \theta + \cos \theta = p\)
On squaring both sides, we get
\((\sin \theta + \cos \theta)^2 = p^2\)
\(\Rightarrow \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = p^2\)
\(\Rightarrow 1 + 2 \sin \theta \cos \theta = p^2\) [\(\because \sin^2 \theta + \cos^2 \theta = 1\)]
\(\Rightarrow 1 + \frac{2p}{q} = p^2\)
\(\Rightarrow q + 2p = p^2q \Rightarrow p^2q - q = 2p\)
\(\Rightarrow q(p^2 - 1) = 2p\). 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: To prove : \(\frac{(1 + \cot \theta + \tan \theta)(\sin \theta - \cos \theta)}{\sec^3 \theta - \csc^3 \theta} = \sin^2 \theta \cos^2 \theta\)
Proof : \(L.H.S. = \frac{\left( 1 + \frac{\cos \theta}{\sin \theta} + \frac{\sin \theta}{\cos \theta} \right)(\sin \theta - \cos \theta)}{\frac{1}{\cos^3 \theta} - \frac{1}{\sin^3 \theta}}\)
\(= \frac{\left( \frac{\sin \theta \cos \theta + \cos^2 \theta + \sin^2 \theta}{\sin \theta \cos \theta} \right)(\sin \theta - \cos \theta)}{\frac{\sin^3 \theta - \cos^3 \theta}{\sin^3 \theta \cos^3 \theta}}\)
\(= \frac{(\sin \theta \cos \theta + 1)(\sin \theta - \cos \theta)}{\sin \theta \cos \theta} \times \frac{\sin^3 \theta \cos^3 \theta}{\sin^3 \theta - \cos^3 \theta}\)
\(= \frac{(\sin \theta \cos \theta + 1)(\sin \theta - \cos \theta)}{\sin \theta \cos \theta} \times \frac{\sin^3 \theta \cos^3 \theta}{(\sin \theta - \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}\)
\([ \because a^3 - b^3 = (a - b)(a^2 + ab + b^2) ]\)
\(= \frac{(1 + \sin \theta \cos \theta)}{(1 + \sin \theta \cos \theta)} \times \cos^2 \theta \sin^2 \theta\)
\(= \sin^2 \theta \cos^2 \theta = RHS\). Hence proved.

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

Question. A moving boat is observed from the top of a 150 m high cliff moving away from it. The angle of depression of the boat changes from \(60^\circ\) to \(45^\circ\) in 2 minutes. Find the speed of the boat in m/min. 
Answer: Let A be the cliff from where the position of the ships are observed and C and D be two positions of ships, which changes from C to D in 2 minutes.
\(\angle EAC = \angle ACB = 60^\circ\)
\(\angle EAD = \angle ADB = 45^\circ\)
\(AB = 150\) m
Now, In \(\Delta ABC\)
\(\tan 60^\circ = \frac{AB}{BC} \Rightarrow \sqrt{3} = \frac{150}{BC} \Rightarrow BC = \frac{150}{\sqrt{3}} = 50\sqrt{3}\) m
Now, in \(\Delta ABD\):
\(\tan 45^\circ = \frac{AB}{BD} \Rightarrow 1 = \frac{150}{BD} \Rightarrow BD = 150\) m
\(\therefore CD = BD - CB = (150 - 50\sqrt{3})\) m
Distance travelled in 2 minutes: \((150 - 50\sqrt{3})\) m
\(\therefore\) Distance travelled in 1 minutes: \(\frac{150 - 50\sqrt{3}}{2}\) m
\(= (75 - 25\sqrt{3})\) m = \(75 - 25 \times 1.732 = 75 - 43.3 = 31.7\) m/min
Hence, the speed of the boat is 31.7 m/min.

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

Question. Prove that : \(\frac{\sin \theta}{\cot \theta + \csc \theta} = 2 + \frac{\sin \theta}{\cot \theta - \csc \theta}\) 
Answer: To prove: \(\frac{\sin \theta}{\cot \theta + \csc \theta} - \frac{\sin \theta}{\cot \theta - \csc \theta} = 2\)
Consider, \(\frac{\sin \theta}{\cot \theta + \csc \theta} - \frac{\sin \theta}{\cot \theta - \csc \theta}\)
\(= \sin \theta \left[ \frac{1}{\cot \theta + \csc \theta} - \frac{1}{\cot \theta - \csc \theta} \right]\)
\(= \sin \theta \left[ \frac{(\cot \theta - \csc \theta) - (\cot \theta + \csc \theta)}{\cot^2 \theta - \csc^2 \theta} \right]\)
\(= \sin \theta \left[ \frac{-2 \csc \theta}{-1} \right]\) [\(\because 1 + \cot^2 \theta = \csc^2 \theta \Rightarrow \cot^2 \theta - \csc^2 \theta = -1\)]
\(= 2 \sin \theta \csc \theta = 2 \sin \theta \times \frac{1}{\sin \theta} = 2\).
Hence, Proved.

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