Question

$\frac{d}{dx}\left\{{\tan }^{-1}\left(\frac{a\sin x+b\cos x}{a\cos x-b\sin x}\right)\right\}=?$

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

Answer 1

Answer: (A)

$0$

$\begin{array}{c}y={\tan }^{-1}\left[\frac{a\cos x+b\sin x}{b\cos x+a\sin x}\right]\\ a=r\sin \theta ,b=r\cos \theta ,weget\\ y={\tan }^{-1}\left[\frac{r(\sin \theta \cos x+\cos \theta \cdot \sin x)}{r(\cos \theta \cdot \cos x-\sin \theta \cdot \sin x)}\right]\\ y={\tan }^{-1}\left|\frac{\sin (\theta +x)}{\cos (x+\theta )}\right|\left[\because \frac{a}{B}=\tan \theta \right]\\ \left[\theta ={\tan }^{-1}\frac{a}{b}\right]\\ y={\tan }^{-1}\left[\tan \theta +x\right]\\ y=\theta +x={\tan }^{-1}\frac{a}{b}+x\\ \frac{dy}{dx}=\frac{d({\tan }^{-1}\frac{a}{b}}{dx}+\frac{d\left(x\right)}{dx}\\ \therefore \frac{dy}{dx}=0+1-1Ans.\end{array}$