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Mathematics Question on Continuity and differentiability

If y=tan⁡−1(sec⁡x−tan⁡x)y = \tan^{-1} ( \sec x - \tan x) , then dydx=\frac{dy}{dx} =

A

22

B

−2-2

C

12 \frac{1}{2}

D

−12 - \frac{1}{2}

Answer

−12 - \frac{1}{2}

Explanation

Solution

y=tan⁡−1(sec⁡x−tan⁡x)y = \tan^{-1} \left(\sec x-\tan x\right)
⇒dydx=1.(sec⁡x × tan⁡x − sec⁡2x)1+(sec⁡x−tan⁡x)2\Rightarrow \frac{dy}{dx} = \frac{1.\left(\sec x \, \times \, \tan x \,-\, \sec^{2} x\right)}{1+\left(\sec x - \tan x\right)^{2}}
=sec⁡x(tan⁡x−sec⁡x)1+sec⁡2x+tan⁡2x−2sec⁡xtan⁡x= \frac{\sec x \left( \tan x -\sec x \right)}{1+\sec^{2} x + \tan^{2} x-2 \sec x \tan x }
=sec⁡x(tan⁡x−sec⁡x)2sec⁡2x−2sec⁡xtan⁡x= \frac{\sec x \left( \tan x - \sec x \right)}{2 \sec^{2} x - 2 \sec x \tan x}
=sec⁡x(tan⁡x−sec⁡x)−2sec⁡x(tan⁡x−sec⁡x)=−12=\frac{\sec x \left(\tan x -\sec x\right)}{-2 \sec x\left(\tan x - \sec x\right)} =\frac{-1}{2}