Solveeit Logo
Login

Question

Mathematics Question on Inverse Trigonometric Functions

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

A

1+x22+x2\sqrt{\frac{1+x^2}{2+x^2}}

B

1−x22+x2\sqrt{\frac{1-x^2}{2+x^2}}

C

1+x22−x2\sqrt{\frac{1+x^2}{2-x^2}}

D

2+x21+x2\sqrt{\frac{2+x^2}{1+x^2}}

Answer

1+x22+x2\sqrt{\frac{1+x^2}{2+x^2}}

Explanation

Solution

cos(tan−1x)=cos θcos\left(tan^{-1} x\right) = cos\, \theta where θ=tan−1x \theta = tan^{-1} x i.e., tan θ=xtan \,\theta = x ∴cosθ=11+x2\therefore cos \theta = \frac{1}{ \sqrt{1+x^{2}}} ∴cos(tan−1x)=11+x2 \therefore cos \left(tan^{-1}x\right) = \frac{1}{ \sqrt{1+x^{2}} } cot−1[cos(tan−1x)]=cot−1[11+x2]=tcot^{-1} \left[ cos\left(tan^{-1}x\right)\right] = cot^{-1} \left[ \frac{1}{\sqrt{1+x^{2}}}\right] = t ⇒cot  t=11+x2\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}}}