Question:
Prove \(tan^{-1}\sqrt{x}=\frac{1}{2}cos^{-1}(\frac{1-x}{1+x}),x∈[0,1]\)
Prove \(tan^{-1}\sqrt{x}=\frac{1}{2}cos^{-1}(\frac{1-x}{1+x}),x∈[0,1]\)
Updated On: Oct 12, 2023
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Solution and Explanation
Let \(x=tan^2θ.\)
Then \(\sqrt{x}=tanθ. =>θ=tan^{-1}\sqrt{x}.\)
so \(\frac {1-x}{1+x}\) = \(\frac{1-tan^2θ}{1+tan^2θ}\) =cos2θ.
Now we have,
RHS=\(\frac12 \cos^{-1}\frac{1-x}{1+x} = \frac12\cos^{-1}(\cos2\theta)= \frac12\times2\theta=\theta=\tan^{-1}\sqrt{x}\)=LHS.
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