Question

Statement I $\tan ^{-1} x+\tan ^{-1} \frac{2 x}{1-x^2}=\tan ^{-1}\left(\frac{3 x-x^3}{1-3 x^2}\right),|x|<\frac{1}{\sqrt{3}}$
Statement II The value of $\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right)$, $|x|>1$ 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{3 x-x^3}{1-3 x^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{2 x}{1-x^2}=\text { LHS }$
II. We have, $\cos \left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right)=\cos \left(\frac{\pi}{2}\right)=0$