Question

If $y={\tan }^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right),\mathrm{then}\frac{dy}{dx}$ is equal to
(a) $\frac{1}{2}$
(b) 0
(c) 1
(d) none of these

Answer 1

(c) 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y={\tan }^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{1}{1+{\left({\displaystyle \frac{\sin x+\cos x}{\cos x-\sin x}}\right)}^2}\frac{d}{dx}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(\cos x-\sin x\right)}^2}{{\left(\cos x-\sin x\right)}^2+{\left(\sin x+\cos x\right)}^2}\left[\frac{\left(\cos x-\sin x\right){\displaystyle \frac{d}{dx}}\left(\sin x+\cos x\right)-\left(\sin x+\cos x\right){\displaystyle \frac{d}{dx}}\left(\cos x-\sin x\right)}{{\left(\cos x-\sin x\right)}^2}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(\cos x-\sin x\right)}^2}{{\left(\cos x-\sin x\right)}^2+{\left(\sin x+\cos x\right)}^2}\left[\frac{\left(\cos x-\sin x\right)\left(\cos x-\sin x\right)-\left(\sin x+\cos x\right)\left(-\sin x-\cos x\right)}{{\left(\cos x-\sin x\right)}^2}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(\cos x-\sin x\right)}^2}{{\left(\cos x-\sin x\right)}^2+{\left(\sin x+\cos x\right)}^2}\left[\frac{\left(\cos x-\sin x\right)\left(\cos x-\sin x\right)+\left(\sin x+\cos x\right)\left(\sin x+\cos x\right)}{{\left(\cos x-\sin x\right)}^2}\right] \\ \Rightarrow \frac{dy}{dx}=\frac{{\left(\cos x-\sin x\right)}^2}{{\left(\cos x-\sin x\right)}^2+{\left(\sin x+\cos x\right)}^2}\times \frac{{\left(\cos x-\sin x\right)}^2+{\left(\sin x+\cos x\right)}^2}{{\left(\cos x-\sin x\right)}^2} \\ \Rightarrow \frac{dy}{dx}=1 \end{gathered}$$