Prove that

    9. \tan^{-1} \sqrt{x} = \frac{1}{2}\cos^{-1}\frac{1-x}{1+x},\;\;x\in [0,1]

Answers (1)

By observing the square root we will first put

x= \tan^2 \theta.

Then,

we have \tan^{-1} \sqrt{\tan^2 \theta} = \frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta}

or, R.H.S.

\frac{1}{2}\cos^{-1}\frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \frac{1}{2}\cos^{-1}(cos2 \theta)

= \frac{1}{2}\times 2\theta = \theta.

L.H.S. \tan^{-1} \sqrt{\tan^2 \theta} = \tan^{-1}(\tan \theta) = \theta

hence L.H.S. = R.H.S proved.

 

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