Question

Show that $tan\left(\frac{1}{2}si{n}^{-1}\frac{3}{4}\right)=\frac{4-\sqrt{7}}{3}$ and justify why the other value $\frac{4+\sqrt{7}}{3}$ is ignored?

Answer 1

We have $tan\left(\frac{1}{2}si{n}^{-1}\frac{3}{4}\right)=\frac{4-\sqrt{7}}{3}$

$\therefore LHS=tan\left[\frac{1}{2}si{n}^{-1}\left(\frac{3}{4}\right)\right]$

$Let\frac{1}{2}si{n}^{-1}\frac{3}{4}=\theta \Rightarrow si{n}^{-1}\frac{3}{4}=2\theta \Rightarrow sin2\theta =\frac{3}{4}\Rightarrow \frac{2tan\theta }{1+ta{n}^2\theta }=\frac{3}{4}\Rightarrow 3+3ta{n}^2\theta =8tan\theta \Rightarrow 3ta{n}^2\theta -8tan\theta +3=0Lettan\theta =y\therefore 3y^2-8y+3=0\Rightarrow y\frac{+8\pm \sqrt{64-4\times 3\times 2}}{2\times 3}=\frac{8\pm \sqrt{28}}{6}=\frac{2[4\pm \sqrt{7}]}{2.3}\Rightarrow tan\theta =\frac{4\pm \sqrt{7}}{3}\Rightarrow \theta =ta{n}^{-1}\left[\frac{4\pm \sqrt{7}}{3}\right]\{buttan\frac{4+\sqrt{7}}{3}>1,sincemax\left[tan\left(\frac{1}{2}si{n}^{-1}\frac{3}{4}\right)\right]=1\}\therefore LHS=tan\left(ta{n}^{-1}\right)\left(\frac{4-\sqrt{7}}{3}\right)=\frac{4-\sqrt{7}}{3}=RHSNoteSince,\frac{\pi }{2}\le si{n}^{-1}\frac{3}{4}\le \pi /2\Rightarrow \frac{-\pi }{4}\le \frac{1}{2}si{n}^{-1}\frac{3}{4}\le \pi /4\therefore tan\left(\frac{-\pi }{4}\right)\le tan\frac{1}{2}\left(si{n}^{-1}\frac{3}{4}\right)\le tan\frac{\pi }{4}\Rightarrow -1\le tan\left(\frac{1}{2}si{n}^{-1}\frac{3}{4}\right)\le 1$