#### Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 38 Maths Textbook Solution.

Answer: $\frac{dy}{dx}=\frac{1}{2},$ Hence,$\frac{dy}{dx}$ is independent of x

Hint:

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

Given:

$y=\cot ^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\}$

Solution:

\begin{aligned} &y=\cot ^{-1}\left\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\} \\ &\text { Then, } \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \\ &\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \times \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \end{aligned}

\begin{aligned} &\frac{(\sqrt{1+\sin x}+\sqrt{1+\sin x})^{2}}{(1+\sin x)-(1-\sin x)} \\ &\begin{array}{l} \text { Using }(a-b)(a+b)=a^{2}-b^{2} \\ (a+b)^{2}=a^{2}+b^{2}+2 a b \\ \frac{(1+\sin x)+(00(1-\sin x)+2 \sqrt{(1-\sin x)(1+\sin x)}}{(1+\sin x)-(1-\sin x)} \\ \frac{1+\sin x+1-\sin x+2 \sqrt{(1-\sin x)(1+\sin x)}}{1+\sin x-1+\sin x} \end{array} \end{aligned}

\begin{aligned} &\Rightarrow \frac{2+2 \sqrt{1+\sin x-\sin x-\sin ^{2} x}}{2 \sin x} \\ &\Rightarrow \frac{2+2 \sqrt{1-\sin ^{2} x}}{2 \sin x} \\ &\Rightarrow \frac{2\left(1+\sqrt{1-\sin ^{2} x}\right)}{2 \sin x} \\ &\Rightarrow \frac{1+\sqrt{1 \sin ^{2} x}}{\sin x} \end{aligned}

Using $\cos ^{2}x+\sin ^{2}x=1$

\begin{aligned} &\Rightarrow \frac{1+\sqrt{\cos ^{2} x}}{\sin x} \\ &\Rightarrow \frac{1+\cos x}{\sin x} \\ &\text { Using } \sin 2 \theta=2 \sin \theta \cos \theta \\ &1+\operatorname{Cos} 2 \theta=2 \cos ^{2} \theta \\ &\Rightarrow \frac{2 \cos ^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \\ &\Rightarrow \cot \frac{x}{2} \end{aligned}                                                        $\left \{ \frac{\cos x}{\sin x}=\cot x\right \}$

Therefore, equation (1) becomes

\begin{aligned} &y=\cot ^{-1}\left(\cot \frac{x}{2}\right) \\ &y=\frac{x}{2} \end{aligned}                                                                                        $\left\{\cot ^{-1}(\cot \theta)=\theta \operatorname{if} \theta^{\varepsilon}\left[\frac{-\pi}{2}, \frac{\mathbf{\pi}}{2}\right]\right\}$

Differentiating it with respect to $x$

$\frac{dy}{dx}=\frac{1}{2}\frac{d}{dx}\left ( x \right )$

$\frac{dy}{dx}=\frac{1}{2}$