Question

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


Chapter name = Inverse Trigonometry Function

Class = 12th

need explanation :)

Answer 1

Let \( x=cos\theta\) \( \Rightarrow \theta = cos^{-1}x\)

\(\sqrt{1+x}=\sqrt{1+cos\theta}=\sqrt{2cos^2\large\frac{\theta}{2}}=\sqrt{2}cos\large\frac{\theta}{2}\)

\(\sqrt{1-x}=\sqrt{1-cos\theta}=\sqrt{2sin^2\large\frac{\theta}{2}}=\sqrt{2}sin\large\frac{\theta}{2}\)

\( \sqrt{1+x}-\sqrt{1-x} \)\(=\sqrt{2}cos\large\frac{\theta}{2}-\sqrt{2}sin\large\frac{\theta}{2} =\) \( \sqrt 2\;(cos\large\frac{\theta}{2}-sin\large\frac{\theta}{2}) \)

\( \sqrt{1+x}+\sqrt{1-x} \)\(=\sqrt{2}cos\large\frac{\theta}{2}+\sqrt{2}sin\large\frac{\theta}{2} \) \(= \sqrt 2\;(cos\large\frac{\theta}{2}+sin\large\frac{\theta}{2}) \)

\(\Large \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}= \large\frac{ \sqrt 2\;(cos\frac{\theta}{2}-sin\large\frac{\theta}{2})}{ \sqrt 2\;(cos\large\frac{\theta}{2}+sin\large\frac{\theta}{2})}\)

\(\frac{cos\frac{\theta}{2}-sin\frac{\theta}{2}}{cos\frac{\theta}{2}+sin\frac{\theta}{2}}\)

 

Dividing numerator and denonimator by  \(cos\large\frac{\theta}{2}\),

 

\( \Rightarrow \)This reduces to \(\;\Large \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}} \) = \(\Large \frac{1-\large\frac{sin\large\frac{\theta}{2}}{cos\large\frac{\theta}{2}}}{1+\large\frac{sin\large\frac{\theta}{2}}{cos\large\frac{\theta}{2}}}\)=\(\Large \frac{1-tan\large\frac{\theta}{2}}{1+tan\large\frac{\theta}{2}}\)

 

Since \(1=tan\large\frac{\pi}{4}\) we can rewrite this as \( \large \frac{tan\large\frac{\pi}{4}-tan\large\frac{\theta}{2}}{1+tan\large\frac{\pi}{4}.tan\large\frac{\theta}{4}}\)

We know that tan(A−B)=tanA−tanB1+tanAtanB

 

By substituting for A=\(\large\frac{\pi}{4}, \:B=\large\frac{\theta}{2}\), we get \(\large \frac{tan\frac{\pi}{4}-tan\large\frac{\theta}{2}}{1+tan\frac{\pi}{4}.tan\large\frac{\theta}{4}}\)\(=tan(\large\frac{\pi}{4}-\large\frac{\theta}{2})\)

\(\Rightarrow\: tan^{-1}tan \bigg( \large\frac{\pi}{4}-\large\frac{\theta}{2} \bigg) = \frac{\pi}{4}-\large\frac{\theta}{2}\)

 

By substituting the value of \(\theta=cos^{-1}x\) \(\Rightarrow\:\large\frac{\pi}{4}-\large\frac{1}{2} cos^{-1}x \) = R.H.S.

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Answer 2

Answer:

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