Q226 / 270 · NDA 2016 (Phase 1) · Mathematics
Consider the following statements:
1. There exists $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ for which $\tan ^{-1}(\tan \theta) \neq \theta$.
2. $\quad \sin ^{-1}\left(\frac{1}{3}\right)-\sin ^{-1}\left(\frac{1}{5}\right)=\sin ^{-1}\left(\frac{2 \sqrt{2}(\sqrt{3}-1)}{15}\right)$
Which of the above statements is/are correct?
- A1 only
- Correct answerB2 only
- CBoth 1 and 2
- DNeither 1 nor 2
Solution
$\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \text { for which } \tan ^{-1}(\tan \theta)=\theta$
Hence, statement (1) is incorrect.
$\begin{aligned}
& \text { if } x \leq 1 ; y \leq 1 \& x^2+y^2 \leq 1 \\
& \therefore \sin ^{-1}(x)-\sin ^{-1} \mathrm{y}=\sin ^{-1}\left[\mathrm{x} \sqrt{1-\mathrm{y}^2}-\mathrm{y} \sqrt{1-\mathrm{x}^2}\right]
\end{aligned}$
$\begin{aligned}
\Rightarrow & \sin ^{-1} \frac{1}{3}-\sin ^{-1} \frac{1}{5} \\
\quad & =\sin ^{-1}\left[\frac{1}{3} \sqrt{1-\frac{1}{25}}-\frac{1}{5} \sqrt{1-\frac{1}{9}}\right] \\
& =\sin ^{-1}\left[\frac{1}{3 \times 5} \sqrt{24}-\frac{1}{5 \times 3} \sqrt{8}\right] \\
& =\sin ^{-1}\left[\frac{\sqrt{8 \times 3}-\sqrt{8}}{15}\right]=\sin ^{-1}\left[\frac{2 \sqrt{2}(\sqrt{3}-1)}{15}\right]
\end{aligned}$
Hence, statement (2) is correct.
Hence, the correct answer is option: 'B'.