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#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 1 Subquestion (iii) Maths Textbook Solution.

$\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )$

Hint:
For solving this, we can use the formula of union trigonometric function,
$\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left ( \frac{a+b}{1-ab} \right )$
Given:
$\tan^{-1}\left ( \frac{1}{4} \right )+\tan^{-1}\left ( \frac{2}{9} \right )=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )$
Explanation:
L.H.S
$= \tan^{-1}\left ( \frac{1}{4} \right )+\tan^{-1}\left ( \frac{2}{9} \right )$
\begin{aligned} &=\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\left(\frac{1}{4} \times \frac{2}{9}\right)}\right) \\ &=\tan ^{-1}\left(\frac{\frac{9+8}{36}}{\frac{36-2}{36}}\right) \\ &=\tan ^{-1}\left(\frac{17}{34}\right) \end{aligned}
$=\tan^{-1}\left ( \frac{1}{2} \right )$
Let’s take   $\tan^{-1}\left ( \frac{1}{2} \right )= \theta$
So,$\tan\left ( \frac{1}{2} \right )= \theta$ [Removing inverse]
$\tan\theta =\left ( \frac{1}{2} \right ) = \frac{Perp}{Base} . \cdot \cdot \cdot \cdot (i)$

Using  the formula of hypotenuse , Hypotenuse =$\sqrt{5}$
$\! \! \! \! \! \! \! \! \sin\theta =\frac{P}{H}=\frac{1}{\sqrt{5}}\\ \theta =\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right ) \cdot \cdot \cdot \cdot (ii)$
Let’s put value of (ii) into (i)
$\! \! \! \! \! \! \! \! \! \tan\left \{ \sin^{-1}\left ( \frac{1}{\sqrt{5}} \right ) \right \}=\frac{1}{2}\\ \tan^{-1}\left ( \frac{1}{2} \right )=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )$                                                                                                   … (iii)
So,  $\tan^{-1}\left ( \frac{1}{2} \right )=\sin^{-1}\left ( \frac{1}{\sqrt{5}} \right )$ =R.H.S
Hence, the prove
Note: We must remember the formula of hypotenuse.