SSC CGL 2024 PYQ — Question 51

The correct answer is option 3 i.e. $x^2-1\over1+x^2$

secθ + tanθ = x   .....(1)

⇒ secθ - tanθ = 1/x   .....(2)

Add equations (1) and (2),

2secθ = x + (1/x)

⇒ secθ = $\frac{(x^2 + 1)}{2x}$⇒ cosθ = $\frac{2x}{(x^2 + 1)}$

⇒ $cos^2\theta=\big[ \frac{2x}{\left(x^2+1 \right)} \big]^2$

⇒ 1 - sin²θ = $\frac{4x^2}{(x^4 + 2x^2 + 1) }$  [∵ cos²θ = 1 - sin²θ]

⇒ sin²θ = 1 - $\frac{4x^2}{(x^4 + 2x^2 + 1) }$ = $\frac{(x^4 - 2x^2 + 1)}{(x^2 + 1)^2}$ = $\frac{(x^2 - 1)^2}{(x^2 + 1)^2}$

⇒ sin θ = $x^2-1\over1+x^2$