Question

$\frac{d}{dx}\left(\tan {}^{-1}(\cos ecx+\cot x)\right)=$

• A. $0$
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{-1}{2}$

Answer 1

The correct option is D $\frac{-1}{2}$

$D=\frac{d}{dx}\left({\tan }^{-1}\right(\csc x+\cot x\left)\right)$

$f\left(x\right)=\frac{d}{dx}\left({\tan }^{-1}g\right(x\left)\right)$

$\frac{d}{dx}\left(g\right(x\left)\right)=\frac{d}{dx}(\csc x+\cot x)$

$=-cscx\cot x-{\csc }^{2}x$

$=-\csc x[\cot x+\csc x]=\frac{-1}{\sin x}\left[\frac{\cos x+1}{\sin x}\right]$

and $\frac{d}{dx}\left({\tan }^{-1}g\right(x\left)\right)$

$D=\left(\frac{1}{1+g(x{)}^2}\right)\frac{d}{dx}\left(g\right(x\left)\right)=\frac{-g\left(x\right)}{\left(\sin x\right)(1+g^2(x)}$

$D\Rightarrow \frac{-[1+\cos x]}{{\sin }^{2}x+(1+\cos x{)}^2}$

$\Rightarrow \frac{-[1+\cos x}{{\sin }^{2}x+1+{\cos }^{2}x+2\cos x}$

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

Option $D$ is correct.