Practicing Success
Practicing Success
CUET Preparation Today
For which value of x, $\sin[\cot^{-1}(x+1)]=\cos(\tan^{-1}x)$. |
$\frac{1}{2}$ 0 1 $-\frac{1}{2}$ |
$-\frac{1}{2}$ |
$\sin[\cot^{-1}(x+1)]=\sin[\sin^{-1}\frac{1}{\sqrt{(x+1)^2+1}}]=\frac{1}{\sqrt{x^2+2x+2}};\cos(\tan^{-1}x)=\cos[\cos^{-1}\frac{1}{\sqrt{1+x^2}}]=\frac{1}{\sqrt{1+x^2}}$ We have $x^2+2x+2=1+x^2$ or $x=-\frac{1}{2}$ |
