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

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

Answers (1)

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}

Posted by

infoexpert21

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