NCERT Class 12 Maths | Chapter 2 Inverse Trigonometric Functions
Question
Show that \[ \tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right) = \frac{4 – \sqrt{7}}{3} \] and justify why the other value \[ \frac{4 + \sqrt{7}}{3} \] is ignored.
[NCERT EXEMPLAR 2.3, Q.18, Page 37]
Solution
We have, \[ \tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right) = \frac{4 – \sqrt{7}}{3} \]
Let \[ \frac{1}{2}\sin^{-1}\frac{3}{4} = \theta \implies \sin^{-1}\frac{3}{4} = 2\theta \]
\[ \sin 2\theta = \frac{3}{4} \implies \frac{2\tan\theta}{1 + \tan^2\theta} = \frac{3}{4} \]
\[ 3 + 3\tan^2\theta = 8\tan\theta \implies 3\tan^2\theta – 8\tan\theta + 3 = 0 \]
Let \(\tan\theta = y\), so the equation becomes: \[ 3y^2 – 8y + 3 = 0 \]
\[ y = \frac{8 \pm \sqrt{64 – 36}}{6} = \frac{8 \pm \sqrt{28}}{6} = \frac{4 \pm \sqrt{7}}{3} \]
\[ \tan\theta = \frac{4 \pm \sqrt{7}}{3} \]
Since \[ \max\left[\tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right)\right] = 1, \] the value \(\frac{4 + \sqrt{7}}{3} > 1\) is ignored.
Therefore, \[ \tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right) = \frac{4 – \sqrt{7}}{3} \]
Note: Since \[ -\frac{\pi}{2} \leq \sin^{-1}\frac{3}{4} \leq \frac{\pi}{2} \implies -\frac{\pi}{4} \leq \frac{1}{2}\sin^{-1}\frac{3}{4} \leq \frac{\pi}{4}, \] we have \[ \tan\left(-\frac{\pi}{4}\right) \leq \tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right) \leq \tan\left(\frac{\pi}{4}\right) \] \[ -1 \leq \tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right) \leq 1. \]