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  • Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 18

Provide solution for RD Sharma maths class 12 chapter Differentiation exercise 10.8 question 18

Answers (1)

Answer: 1

Hint: \text { Let } x=\cos \theta
 

Given: \sin ^{-1} \sqrt{1-x^{2}} \text { w.r.t } \cot ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right)

Explanation:

\text { Let } x=\cos \theta

\begin{aligned} &u=\sin ^{-1} \sqrt{1-x^{2}} \\\\ &=\sin ^{-1} \sqrt{1-\cos ^{2} \theta} \\\\ &=\sin ^{-1}(\sin \theta) \end{aligned}

\begin{aligned} &v=\cot ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \\\\ &=\cot ^{-1}\left(\frac{\cos \theta}{\sqrt{1-\cos ^{2} \theta}}\right) \end{aligned}

=\cot ^{-1}\left(\frac{\cos \theta}{\sin \theta}\right)=\cot ^{-1}(\cot \theta)

Now

\begin{aligned} &0<x<1 \\\\ &0<\cos \theta<1 \\\\ &0<\theta<\frac{\pi}{2} \end{aligned}

\begin{array}{ll} u=\sin ^{-1}(\sin \theta)=\theta & \quad \theta \in\left(0, \frac{\pi}{2}\right) \\\\ v=\cot ^{-1}(\cot \theta)=\theta & 2 \theta \in\left(0, \frac{\pi}{2}\right) \end{array}

\begin{aligned} &u=\cos ^{-1} x \quad\left[\begin{array}{l} x=\cos \theta \\ \theta=\cos ^{-1} x \end{array}\right] \\ &\frac{d u}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}

\begin{aligned} &v=\cos ^{-1} x \\\\ &\frac{d v}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \end{aligned}

\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}=\frac{\frac{-1}{\sqrt{1-x^{2}}}}{\frac{-1}{\sqrt{1-x^{2}}}}=1

\frac{d u}{d v}=1

 

Posted by

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