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SSC CHSL 2025 PYQ — Question 73

SSC CHSL - 12 Nov 2025 - Shift 2 • Quantitative Aptitude • Q73

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1+tan⁡2x1+cot⁡2x×sec⁡x−tan⁡xsec⁡x+tan⁡x=?\frac{1+\tan^2 x}{1+\cot^2 x} \times \frac{\sec x - \tan x}{\sec x + \tan x} = ?

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Correct answer: D

Explanation

1+tan⁡2x1+cot⁡2x×sec⁡x−tan⁡xsec⁡x+tan⁡x=sec⁡2xcsc⁡2x×sec⁡x−tan⁡xsec⁡x+tan⁡x=sin⁡2xcos⁡2x×1−sin⁡x1+sin⁡x=sin⁡2x(1−sin⁡x)(1+sin⁡x)×1−sin⁡x1+sin⁡x=sin⁡2x(1+sin⁡x)2=tan⁡2x(1−sin⁡x)1+sin⁡x\frac{1+\tan^2 x}{1+\cot^2 x} \times \frac{\sec x - \tan x}{\sec x + \tan x}\\ = \frac{\sec^2 x}{\csc^2 x} \times \frac{\sec x - \tan x}{\sec x + \tan x}\\ = \frac{\sin^2 x}{\cos^2 x} \times \frac{1 - \sin x}{1 + \sin x}\\ = \frac{\sin^2 x}{(1 - \sin x)(1 + \sin x)} \times \frac{1 - \sin x}{1 + \sin x}\\ = \frac{\sin^2 x}{(1 + \sin x)^2}\\ = \frac{\tan^2 x (1 - \sin x)}{1 + \sin x}