Question

$\frac{d}{d\theta }\left({\tan }^{-1}\left(\frac{1-\cos \theta }{\sin \theta }\right)\right)$ equals if $-\pi <\theta <\pi$

• A. $\frac{1}{2}$
• B. $1$
• C. $\sec \theta$
• D. $cosec\theta$

Answer 1

Answer: (A)

$\frac{1}{2}$

$\frac{1-\cos \left(x\right)}{\sin \left(x\right)}$

Let $u=\frac{x}{2}$

$=\frac{-(1-2{\sin }^2\left(u\right))+1}{2\cos \left(u\right)\sin \left(u\right)}$

$=\frac{\sin \left(u\right)}{\cos \left(u\right)}$

$=\tan \left(u\right)=tan\left(\frac{x}{2}\right)$

So,

$\frac{d}{d\theta }\left(\arctan \left(\frac{1-\cos \left(\theta \right)}{\sin \left(\theta \right)}\right)\right)=\frac{d}{d\theta }\left(\arctan \left(tan\frac{\theta }{2}\right)\right)=\frac{d}{d\theta }\left(\frac{\theta }{2}\right)=\frac{1}{2}$