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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.

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Solving LHS,

$=\tan\left ( \frac{1}{2} \sin^{-1}\frac{3}{4} \right )$

$Let \frac{1}{2} \sin^{-1}\frac{3}{4} =\theta$

$\Rightarrow \sin^{-1}\frac{3}{4} =2\theta$

$\Rightarrow \frac{3}{4} =\sin2\theta$

$\Rightarrow \sin2\theta= \frac{3}{4}$

$\Rightarrow \frac{2 \tan \theta}{1+\tan^{2}\theta}= \frac{3}{4}$

$\Rightarrow 3+3 \tan^{2}\theta = 8 \tan \theta$

$\Rightarrow 3 \tan^{2}\theta - 8 \tan \theta =3$

$Let \tan \theta=y$

$\therefore 3y^{2}+8y+3=0$

$\Rightarrow y= \frac{8\pm \sqrt{64-4\times 3\times 3}}{2\times 3}$ $\Rightarrow = \frac{8\pm \sqrt{28}}{6}$

$\Rightarrow y=\frac{2\left ( 4\pm \sqrt{7} \right )}{2\times 3}$

$\Rightarrow \tan \theta=\frac{\left ( 4\pm \sqrt{7} \right )}{3}$

$\Rightarrow \theta=\tan^{-1}\frac{\left ( 4\pm \sqrt{7} \right )}{3}$

{ but as we can see , $\frac{ 4+ \sqrt{7} }{3}> 1$ , since $max\left [ \tan\left ( \frac{1}{2} \sin^{-1}\frac{3}{4}\right ) \right ]=1$ }

$\tan\left ( \frac{1}{2} \sin^{-1}\frac{3}{4}\right ) =\frac{4-\sqrt{7}}{3}=RHS$

Note: Scince $-\frac{\pi}{2}\leq sin^{-1}\frac{3}{4}\leq \frac{\pi}{2}$

$\Rightarrow -\frac{\pi}{4}\leq \frac{1}{2}sin^{-1}\frac{3}{4}\leq \frac{\pi}{4}$

$\therefore \tan\left ( -\frac{\pi}{4} \right )\leq \tan\left ( \frac{1}{2}sin^{-1}\frac{3}{4} \right )\leq \tan\left ( \frac{\pi}{4} \right )$

$\Rightarrow -1\leq \tan\left ( \frac{1}{2}\sin^{-1}\frac{3}{4} \right )\leq 1$

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infoexpert24

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