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

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

Answers (1)

Answer: \frac{1}{2} \cos ^{-1} \frac{x}{a}

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

Given:  \tan ^{-1} \sqrt{\frac{a-x}{a+x}},-a<x<a

Explanation:

Let         x=a \cos \theta

then        \theta=\cos ^{-1} \frac{x}{a}

Now,

            \tan ^{-1} \sqrt{\frac{a-x}{a+x}}=\tan ^{-1} \sqrt{\frac{a-a \cos \theta}{a+a \cos \theta}}

            \begin{aligned} &\tan ^{-1} \sqrt{\frac{a(1-\cos \theta)}{a(1+\cos \theta)}} \\ &\tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \end{aligned}

            \tan ^{-1} \sqrt{\frac{2 \sin \frac{2}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \quad\left[\begin{array}{l} \left.\because \begin{array}{l} 1+\cos x=2 \cos ^{2}\left(\frac{x}{2}\right) \\ 1-\cos x=2 \sin ^{2}\left(\frac{x}{2}\right) \end{array}\right] \end{array}\right.

            \tan ^{-1} \sqrt{\tan ^{2} \frac{\theta}{2}} \quad \because \frac{\sin \theta}{\cos \theta}=\tan \theta

            \tan ^{-1}\left(\tan \frac{\theta}{2}\right)

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

            \begin{aligned} &\Rightarrow \quad \frac{\theta}{2} \\ &\Rightarrow \quad \frac{1}{2} \cos ^{-1} \frac{x}{a} \end{aligned}

 

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