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  4. The derivative of tan-1 ((sin x - cos x/sin x + cos x)) with respect to (x/2), where (x ∈ (0, (π/2))) is

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Q. The derivative of $tan^{-1} (\frac{sin\,x - cos\,x}{sin\,x + cos\,x})$ with respect to $\frac{x}{2}$, where $(x \in (0, \frac{\pi}{2}))$ is

Continuity and Differentiability

Solution:

$f\left(x\right) = tan^{-1} \left(\frac{tan \,x -1}{tan\, x +1}\right)$
$ = -tan^{-1}\left(tan\left(\frac{\pi}{4}-x\right)\right)\left[\because \frac{\pi}{4}-x\in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\right] $
So, $f\left(x\right) = -\left(\frac{\pi}{4}-x\right) = x - \frac{\pi}{4}$
Let $y = \frac{x}{2}$
$\Rightarrow f\left(y\right) = 2y - \frac{\pi}{4} $
Now, differentiate w.r.t. $ y, \frac{df(y)}{dy} = 2$.