#### Explain Solution R.D.Sharma Class 12 Chapter 10  Differentiation  Exercise 10.3 Question 32 Maths Textbook Solution.

Answer: $\frac{dy}{dx}=1$

Hint:

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constan } \mathrm{t})=0 ; \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}

Given:

$\tan ^{-1}\left[\frac{\cos x+\sin x}{\cos x-\sin x}\right]$

Solution:

\begin{aligned} &y=\tan ^{-1}\left[\frac{\cos x+\sin x}{\cos x-\sin x}\right] \\ &y=\tan ^{-1}\left[\frac{\frac{\cos x+\sin x}{\cos x}}{\frac{\cos x-\sin x}{\cos x}}\right] \\ &y=\tan ^{-1}\left[\frac{1+\tan x}{1-\tan x}\right] \\ &\text { since, } \frac{\sin x}{\cos x}=\operatorname{tian} x \end{aligned}

\begin{aligned} &y=\tan ^{-1}\left[\frac{\tan \frac{\pi}{4}+\tan x}{1-\frac{\tan \pi}{4} \tan x}\right] \\ &y=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+x\right)\right] \\ &\text { Since } \tan (A+B)=\frac{\tan A+\tan B}{1-\tan \tan B} \\ &y=\frac{\pi}{4}+x \end{aligned}

Differentiating it with respect to x,

$\frac{dy}{dx}=0+1$                                                                                        $\left\{\begin{array}{l} \frac{\mathrm{d}}{\mathrm{dx}}(\text { constant })=0 \\ \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1 \end{array}\right\}$

$\frac{dy}{dx}=1$