Question

Statement I ${\tan }^{-1}x+{\tan }^{-1}\frac{2x}{1-x^2}={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),|x|<\frac{1}{\sqrt{3}}$
Statement II The value of $\cos \left({\sec }^{-1}x+{\mathrm{cosec}}^{-1}x\right)$, $|x|>1$ is 1 .

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

Answer: (A)

I. Let $x=\tan \theta$. Then, $\theta ={\tan }^{-1}x$. We have,
$RHS={\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)={\tan }^{-1}\left(\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }\right)$
$={\tan }^{-1}(\tan 3\theta )=3\theta =3{\tan }^{-1}x$
$={\tan }^{-1}x+2{\tan }^{-1}x$
$={\tan }^{-1}x+{\tan }^{-1}\frac{2x}{1-x^2}=\text{ LHS }$
II. We have, $\cos \left({\sec }^{-1}x+{\mathrm{cosec}}^{-1}x\right)=\cos \left(\frac{\pi }{2}\right)=0$