Question

differentiate tan-1(1+cosx/sinx) with respect to x.

Answer 1

hope this help........

Answer 2

Answer:

$d(\tan^{-1}(\frac{1+\cos x}{\sin x}))=-\frac{1}{2}$

Step-by-step explanation:

$\text{Let y = }\tan^{-1}(\frac{1+\cos x}{\sin x})\\\\Using, 1+\cos\theta=2\cos^2 \frac{\theta}{2}\text{ and }\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\\\\implies y=\tan^{-1}(\frac{2\cos^2 \frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}})\\\\\implies y=\tan^{-1}(\frac{2\cos\frac{\theta}{2}}{2\sin\frac{\theta}{2}})\\\\\implies y=\tan^{-1}(\cot\frac{\theta}{2})\\\\\implies y = \tan^{-1}(\tan(\frac{\pi}{2}-\frac{\theta}{2}))\\\\\implies y=\frac{\pi}{2}-\frac{\theta}{2}$

Now, differentiating this above equation w.r.t. x

$\frac{dy}{dx}=0-\frac{1}{2}\\\\\implies d(\tan^{-1}(\frac{1+\cos x}{\sin x}))=-\frac{1}{2}$