Q66 / 180 · KCET 2007 · Mathematics
The solution of $\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}$ is
- A$-\frac{1}{\sqrt{3}}$
- B$\frac{1}{\sqrt{3}}$
- C$-\sqrt{3}$
- Correct answerD$\sqrt{3}$
Solution
We have, $\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}$
$$
\Rightarrow \quad \tan ^{-1} x+2 \tan ^{-1} \frac{1}{x}=\frac{2 \pi}{3}
$$
$$
\Rightarrow \tan ^{-1} x+\tan ^{-1}\left(\frac{2\left(\frac{1}{x}\right)}{1-\left(\frac{1}{x}\right)^{2}}\right)=\frac{2 \pi}{3}
$$
$\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]$
$\Rightarrow \quad \tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{x^{2}-1}\right)=\frac{2 \pi}{3}$
$\Rightarrow \quad \tan ^{-1}\left(\frac{x+\frac{2 x}{x^{2}-1}}{1-\frac{2 x^{2}}{x^{2}-1}}\right)=\frac{2 \pi}{3}$
$\Rightarrow \quad \frac{x^{3}-x+2 x}{x^{2}-1-2 x^{2}}=\tan \left(\frac{2 \pi}{3}\right)$
$\Rightarrow \quad \frac{x^{3}+x}{-1-x^{2}}=\tan \left(\frac{2 \pi}{3}\right)$
$\Rightarrow \quad \frac{x\left(x^{2}+1\right)}{-1\left(x^{2}+1\right)}=-\sqrt{3}$
$\Rightarrow \quad x=\sqrt{3}$
Hence, the correct answer is option: 'D'.