Question

$cos[tan^{-1}\\{sin(cot^{-1}x)\\}]$ is equal to

• A. $\sqrt{\frac{x^{2}+2}{x^{2}+3}}$
• B. $\sqrt{\frac{x^{2}+2}{x^{2}+1}}$
• C. $\sqrt{\frac{x^{2}+1}{x^{2}+2}}$
• D. None of these

Answer 1

Answer: (C)

$\sqrt{\frac{x^{2}+1}{x^{2}+2}}$

Answer 2

We have, $cos \,[tan^{-1}\\{sin(cot^{-1}x)\\}]$ Let $cot^{-1} \,x = \theta \Rightarrow cot\, \theta = x$ $\Rightarrow sin\,\theta = \frac{1}{\sqrt{1+x^{2}}}$ \therefore cos\left[tan^{-1}\left\\{sin\,\theta\right\\}\right] = cos \left[tan^{-1}\left(\frac{1}{\sqrt{1+x^{2}}}\right)\right] Again, let $tan^{-1} \frac{1}{\sqrt{1+x^{2}}} = \phi$ $\Rightarrow tan\,\phi= \frac{1}{\sqrt{1+x^{2}}}$ $\Rightarrow cos\,\phi = \frac{\sqrt{1+x^{2}}}{\sqrt{2+x^{2}}}$ $\therefore cos \left[tan^{-1} \frac{1}{\sqrt{1+x^{2}}}\right] = cos\,\phi$ $= \sqrt{\frac{x^{2}+1}{x^{2}+2}}$