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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions4 Marks
Question
Show that $2 \sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{24}{7}\right)$.
✓

Answer

Let $\sin ^2\left(\frac{3}{5}\right)=x$.
Then $\sin x=\frac{3}{5}$, where $0<x<\frac{\pi}{2}$
$\therefore \cos x>0$
Now, $\cos x=\sqrt{1-\sin ^2 x}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
$\therefore \tan x=\frac{\sin x}{\cos x}=\frac{(3 / 5)}{(4 / 5)}=\frac{3}{4}$
$\therefore x=\tan ^{-1}\left(\frac{3}{4}\right)$
$\therefore \sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{3}{4}\right)$
Now, LHS $=2 \sin ^{-1}\left(\frac{3}{5}\right)=2 \tan ^{-1}\left(\frac{3}{4}\right)$
$=\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{3}{4}\right)$
$=\tan ^{-1}\left[\frac{\frac{3}{4}+\frac{3}{4}}{1-\frac{3}{4} \times \frac{3}{4}}\right]=\tan ^{-1}\left[\frac{12+12}{16-9}\right]$
$=\tan ^{-1}\left(\frac{24}{7}\right)=$ RHS.
Alternative Method:
$\mathrm{LHS}=2 \sin ^{-1}\left(\frac{3}{5}\right)=2 \tan ^{-1}\left(\frac{3}{4}\right)$
$=\tan ^{-1}\left[\frac{2\left(\frac{3}{4}\right)}{1-\left(\frac{3}{4}\right)^2}\right]$
$\cdots\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]$
$=\tan ^{-1}\left[\frac{\left(\frac{3}{2}\right)}{1-\left(\frac{9}{16}\right)}\right]$
$=\tan ^{-1}\left(\frac{3}{2} \times \frac{16}{7}\right)$
$\therefore \tan ^{-1}\left(\frac{24}{7}\right)=$ RHS

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