Question

Prove that cos bracket tan inverse sin cot inverse x equal to x square + 1 by x square + 2 raise to 1 by 2

Answer 1

Answer:

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Answer 2

To prove : $\cos [\tan^-^1{\sin(\cot^-^1x)}]= \sqrt{\frac{1+x^2}{2+x^2}}$

Explanation:

solving LHS

let x = $\cot \theta$

then

$\cos [\tan^-^1{\sin(\cot^-^1x)}]\\\\cos [tan^-^1(\sin\theta)]\\\\\cos\left [ \tan^-^1\left ( \frac{1}{\sqrt{1+\cot^2\theta}} \right ) \right ]\\\\\cos\left [ \tan^-^1\left ( \frac{1}{\sqrt{1+\x^2}} \right ) \right ]..(1)\\\\let \\\\\theta_1 = \tan^-^1( \frac{1}{\sqrt{1+ x^2}} \right )\\\\\tan\theta_1 = \frac{1}{\sqrt{1+x^2}}$

$\cos\theta_1 = \frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}\\\\\theta_1 =\cos^-^1= \frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}\\\\\text{from 1 }\\\\\cos \left [ \cos^-^1 \frac{\sqrt{1+x^2}}{\sqrt{2+x^2}} \right ]= \frac{\sqrt{1+x^2}}{\sqrt{2+x^2}}$

hence proved

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