Question

Find derivative of ${\cot }^{-1}(\sec x+\tan x)$.

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $2$
• D. $-2$

Answer 1

Answer: (B)

$-\frac{1}{2}$

$\sec x+\tan x=\frac{1}{\cos x}+\frac{\sin x}{\cos x}$

$=\frac{1+\sin x}{\cos x}$

$=\frac{{\sin }^2\frac{x}{2}+{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}{\cos x}$ $[{\sin }^{2}a+{\cos }^{2}a=1\sin \left(2a\right)=2\sin a\cos a]$

$=\frac{{(\sin \frac{x}{2}+\cos \frac{x}{2})}^2}{{\cos }^2\frac{x}{2}-{\sin }^2\frac{x}{2}}$ $\left[\cos \right(2a)={\cos }^{2}a-{\sin }^{2}a]$

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

$=\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}$

$=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}$

$=\frac{\tan \frac{\pi }{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi }{4}.\tan \frac{x}{2}}$

$=\tan (\frac{\pi }{4}+\frac{x}{2})$

$=\cot [\frac{\pi }{2}-(\frac{\pi }{4}+\frac{x}{2})]$

$=\cot (\frac{\pi }{4}-\frac{x}{2})$

Hence, $f\left(x\right)={\cot }^{-1}(\sec x+\tan x)={\cot }^{-1}\left(\cot (\frac{\pi }{4}-\frac{x}{2})\right)$

$=\frac{\pi }{4}-\frac{x}{2}$

By the first principle of differentiation,

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{(\frac{\pi }{4}-\frac{x+h}{2})-(\frac{\pi }{4}-\frac{x}{2})}{h}$

$=\underset{h\to 0}{lim}\frac{-\frac{h}{2}}{h}$

$=-\frac{1}{2}$