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

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

Answers (1)

Answer:
\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.                                                                                

Posted by

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