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provide solution for RD Sharma maths class 12 chapter Inverse Trignometric Functions exercise 3.7 question 7 sub question (x)

Answers (1)

Answer:  \sqrt{1-x^{2}}

Hint:  The range of principal value of \sin ^{-1} is \left [ -\frac{\pi }{2} ,\frac{\pi }{2}\right ]

Given:  \sin \left[2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right]

Explanation:

First we solve  2 \tan ^{-} \sqrt{\frac{1-x}{1+x}}

Let            x=\cos 2 A

Therefore,

            2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}=2 \tan ^{-1} \sqrt{\frac{1-\cos 2 A}{1+\cos 2 A}}

            \left[1-\cos 2 x=2 \sin ^{2} x \& 1+\cos 2 A=2 \cos ^{2} A\right]

            =2 \tan ^{-1} \sqrt{\frac{2 \sin ^{2} A}{2 \cos ^{2} A}}

            \begin{aligned} &=2 \tan ^{-1} \sqrt{\tan ^{2} A} \\ &=2 \tan ^{-1}(\tan A) \end{aligned}

As we know,  \tan ^{-1}(\tan x)=x, x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

            =2 \tan ^{-1}(\tan A)=2 A

Therefore,

            \sin \left[2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right]

            =\sin 2 A \quad \therefore\left(1-\sin ^{2} \theta=\cos ^{2} \theta\right)

            =\sqrt{1-\cos ^{2} 2 A}

Now put value of \cos 2 A=x

            \sqrt{1-\cos ^{2} 2 A}=\sqrt{1-x^{2}}

 

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