Question

$cos\left[ta{n}^{-1}\right(sin\left(co{t}^{-1}x\right)\left)\right]$ 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. $\sqrt{\frac{x^2-1}{x^2+1}}$

Answer 1

The correct option is B $\sqrt{\frac{x^2+2}{x^2+1}}$

Let ${\text{cot}}^{-1}x=y⟹x=\cot y⟹\sin y=\frac{1}{\sqrt{1+x^2}}$

Let ${\text{tan}}^{-1}\frac{1}{\sqrt{x^2+1}}=z⟹\tan z=\frac{1}{\sqrt{1+x^2}}$

So $\cos z=\frac{1}{\sqrt{1+(\frac{1}{\sqrt{x^2+1}}{)}^2}}=\frac{1}{\sqrt{1+\frac{1}{x^2+1}}}=\frac{1}{\sqrt{\frac{x^2+2}{x^2+1}}}=\sqrt{\frac{x^2+1}{x^2+2}}$

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