SSC CHSL 12 Nov 2025 Shift 2 Paper

Concept:Use trigonometric identities to simplify each fraction separately, then multiply them.Explanation:First, simplify $\frac{1+\tan^2 x}{1+\cot^2 x}$.Using $1+\tan^2 x = \sec^2 x$ and $1+\cot^2 x = \csc^2 x$, we get $\frac{\sec^2 x}{\csc^2 x}$.Since $\sec x = \frac{1}{\cos x}$ and $\csc x = \frac{1}{\sin x}$, this becomes $\frac{1/\cos^2 x}{1/\sin^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$.Next, simplify $\frac{\sec x - \tan x}{\sec x + \tan x}$.Multiply numerator and denominator by $(\sec x - \tan x)$:Numerator becomes $(\sec x - \tan x)^2$, denominator becomes $\sec^2 x - \tan^2 x = 1$.So the fraction is $(\sec x - \tan x)^2$.Write in terms of $\sin$ and $\cos$: $\left(\frac{1}{\cos x} - \frac{\sin x}{\cos x}\right)^2 = \frac{(1-\sin x)^2}{\cos^2 x}$.Using $\cos^2 x = 1 - \sin^2 x = (1-\sin x)(1+\sin x)$, we get $\frac{(1-\sin x)^2}{(1-\sin x)(1+\sin x)} = \frac{1-\sin x}{1+\sin x}$.Now multiply the two simplified parts: $\tan^2 x \times \frac{1-\sin x}{1+\sin x}$.Answer:$\frac{\tan^2 x (1-\sin x)}{1+\sin x}$ (Option D)