Show that tan-1 (1/2 sin-1 3/4) = (4 – √7)/ 3 and justify why the other value (4 + √7)/ 3 is ignored.
We have, tan-1 (1/2 sin-1 3/4)
Let ½ sin-1 ¾ = θ ⇒ sin-1 ¾ = 2θ ⇒ sin 2θ = ¾
2 tan θ/ 1 + tan2 θ = ¾
3 tan2 θ – 8 tan θ + 3 = 0
$\tan \theta=\frac{8 \pm \sqrt{64-36}}{6}$
$\tan \theta=\frac{8 \pm \sqrt{28}}{6}=\frac{8 \pm 2 \sqrt{7}}{6}=\frac{4 \pm \sqrt{7}}{3}$
Now,
$-\frac{\pi}{2} \leq \sin ^{-1} \frac{3}{4} \leq \frac{\pi}{2}$
$\frac{-\pi}{4} \leq \frac{1}{2} \sin ^{-1} \frac{3}{4} \leq \frac{\pi}{2}$
$\tan \left(\frac{-\pi}{4}\right) \leq \tan \left(\frac{1}{2}\left(\sin ^{-1} \frac{3}{4}\right)\right) \leq \tan \frac{\pi}{4}$
$-1 \leq \tan \left(\frac{1}{2} \sin ^{-1} \frac{3}{4}\right) \leq 1$
Hence,
$\tan \theta=\frac{4-\sqrt{7}}{3}\left(\tan \theta=\frac{4+\sqrt{7}}{3}>1\right.$, which is not possible $)$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.