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Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsTrigonometric Functions2 Marks
MCQ
If $\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\sin ^{-1}\left(\frac{2 b}{1+b^2}\right)=2 \tan ^{-1} x$, then $x=$
  • A
    $\frac{a-b}{1+a b}$
  • B
    $\frac{b}{1+a b}$
  • C
    $\frac{ b }{1- ab }$
  • ✓
    $\frac{a+b}{1-a b}$
✓

Answer

Correct option: D.
$\frac{a+b}{1-a b}$
(D) $\sin ^{-1}\left(\frac{2 a}{1+a^2}\right)+\sin ^{-1}\left(\frac{2 b}{1+b^2}\right)=2 \tan ^{-1} x$
Putting $a =\tan \theta$ and $b =\tan \phi$, we get
$\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)+\sin ^{-1}\left(\frac{2 \tan \phi}{1+\tan ^2 \phi}\right)=2 \tan ^{-1} x$
$\Rightarrow \sin ^{-1}[\sin (2 \theta)]+\sin ^{-1}[\sin (2 \phi)]=2 \tan ^{-1} x$
$\begin{array}{l}\Rightarrow 2(\theta+\phi)=2 \tan ^{-1} x \\ \Rightarrow x=\tan (\theta+\phi)\end{array}$
$\Rightarrow x=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}$
Resubstituting the values of $a$ and $b$, we get
$x=\frac{ a + b }{1- ab }$

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