Q66 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
If $\sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$, then $\tan ^{-1} x+\tan ^{-1}\left(\frac{x}{x+1}\right)=$
- A$\frac{\pi}{6}$
- Correct answerB$\frac{\pi}{4}$
- C$\frac{\pi}{3}$
- D$\frac{\pi}{2}$
Solution
$\begin{aligned}
& \sin ^{-1} x-\cos ^{-1} 2 x=\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)-\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right) \\
= & \frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \Rightarrow \sin ^{-1} x=\frac{\pi}{6}+\cos ^{-1} 2 x
\end{aligned}$
Take sine on both sides $x=\frac{1}{2} \times 2 x+\frac{\sqrt{3}}{2} \sqrt{1-4 x^2} \Rightarrow 1-4 x^2=0 \Rightarrow x= \pm \frac{1}{2}$ $y=\frac{1}{2}$ satisfies the given equation. $\begin{aligned} & \tan ^{-1} x+\tan ^{-1}\left(\frac{x}{x+1}\right)=\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\frac{1}{2}}{\frac{1}{2}+1}\right) \\ & =\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan ^{-1} 1=\frac{\pi}{4} . \end{aligned}$
Take sine on both sides $x=\frac{1}{2} \times 2 x+\frac{\sqrt{3}}{2} \sqrt{1-4 x^2} \Rightarrow 1-4 x^2=0 \Rightarrow x= \pm \frac{1}{2}$ $y=\frac{1}{2}$ satisfies the given equation. $\begin{aligned} & \tan ^{-1} x+\tan ^{-1}\left(\frac{x}{x+1}\right)=\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\frac{1}{2}}{\frac{1}{2}+1}\right) \\ & =\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\right)=\tan ^{-1} 1=\frac{\pi}{4} . \end{aligned}$
Hence, the correct answer is option: 'B'.