CBSE Class 12 Mathematics 2023: question 27(a)

Hints

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Step-by-step solution

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1. Put $x=\sin \theta$ (taking $0<x<\frac{1}{\sqrt{2}}$, so $0<\theta <\frac{\pi }{4}$).

   $U={\sec }^{-1}\left(\frac{1}{\sqrt{1-{\sin }^2\theta }}\right)={\sec }^{-1}\left(\frac{1}{\cos \theta }\right)={\sec }^{-1}(\sec \theta )=\theta ={\sin }^{-1}x$

   So $\frac{dU}{dx}=\frac{1}{\sqrt{1-x^2}}$

   1 mark

2. $V={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right)={\sin }^{-1}(2\sin \theta \cos \theta )={\sin }^{-1}(\sin 2\theta )=2\theta =2{\sin }^{-1}x$

   So $\frac{dV}{dx}=\frac{2}{\sqrt{1-x^2}}$

   1 mark

3. $\frac{dU}{dV}=\frac{dU/dx}{dV/dx}=\frac{1/\sqrt{1-x^2}}{2/\sqrt{1-x^2}}=\frac{1}{2}$

   1 mark

Answer: $\frac{dU}{dV}=\frac{1}{2}$

Revise the chapter

• Chapter: Inverse Trigonometric Functions (notes, formulas, where marks are lost and every past question)