Question

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

• A. $\sqrt{\frac{1+x^2}{2+x^2}}$
• B. $\sqrt{\frac{1-x^2}{2+x^2}}$
• C. $\sqrt{\frac{1+x^2}{2-x^2}}$
• D. $\sqrt{\frac{2+x^2}{1+x^2}}$

Answer 1

Answer: (A)

$cos\left(tan^{-1} x\right) = cos\, \theta $
where $ \theta = tan^{-1} x$ i.e., $ tan \,\theta = x $

$ \therefore cos \theta = \frac{1}{ \sqrt{1+x^{2}}} $
$ \therefore cos \left(tan^{-1}x\right) = \frac{1}{ \sqrt{1+x^{2}} }$
$ cot^{-1} \left[ cos\left(tan^{-1}x\right)\right] = cot^{-1} \left[ \frac{1}{\sqrt{1+x^{2}}}\right] = t $
$ \Rightarrow cot\,\, t = \frac{1}{ \sqrt{1+x^{2}}} $

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