Aspire Faculty ID #15291 · Topic: JAMIA MCA 2016 · Just now
JAMIA MCA 2016

If $y = \tan^{-1}\!\left(\dfrac{1 + \tan x}{1 - \tan x}\right)$, then $\dfrac{dy}{dx}$ is equal to …

Solution

We know $\tan(2x) = \dfrac{2\tan x}{1 - \tan^2 x}$. Here, $\dfrac{1 + \tan x}{1 - \tan x} = \tan\!\left(\dfrac{\pi}{4} + x\right)$. So, $y = \tan^{-1}(\tan(\dfrac{\pi}{4} + x)) = \dfrac{\pi}{4} + x$. Hence, $\dfrac{dy}{dx} = 1$.
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