Question

$sin\{ta{n}^{-1}\left(\frac{1-x^2}{2x}\right)+co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)\}$ is equal to

• A. $0$
• B. $1$
• C. $\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is B $1$

$sin\left[ta{n}^{-1}\left(\frac{1-ta{n}^2\theta }{2tan\theta }\right)\right]+co{s}^{-1}\left(\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }\right)Puttingx=tan\theta weget,sin\left[ta{n}^{-1}\left(\frac{1-ta{n}^2\theta }{2tan\theta }\right)\right]+co{s}^{-1}\left(\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }\right)=sin\left[ta{n}^{-1}\right(cot2\theta )+co{s}^{-1}(cos2\theta \left)\right]=sin\left[ta{n}^{-1}tan\right(\pi /2-2\theta )+co{s}^{-1}cos2\theta ]=sin\frac{\pi }{2}=1.$