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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.12 Question 2 Subquestion (i) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.12 Question 2 Subquestion (i) Maths Textbook Solution

Answers (1)

To Prove: \sin^{-1}\frac{63}{65}= \sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}
Hint:  Using R.H.S term, here first we will we will convert \sin^{-1}x and \cos^{-1}x then use formula.
\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}} +y\sqrt{1-x^{2}}\right ]
Solution: Taking R.H.S
R.H.S = \sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}
          == \sin^{-1}\frac{5}{13}+\sin^{-1}\frac{4}{5} \; \; \; \; \; \; \; \; \; [\because \cos^{-1}x=\sin^{-1}\sqrt{1-x^{2}}]
Then we will use the formula
\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}} +y\sqrt{1-x^{2}}\right ]
\begin{aligned} &= \sin ^{-1}\left[\frac{5}{13} \sqrt{1-\left(\frac{4}{5}\right)^{2}}+\frac{4}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}\right] \\ &= \sin ^{-1}\left(\frac{5}{13} \times \frac{3}{5}+\frac{4}{5} \times \frac{12}{13}\right) \\ &= \sin ^{-1}\left(\frac{15}{65}+\frac{48}{65}\right) \end{aligned}
 = \sin^{-1}\left ( \frac{63}{65} \right )
= L.H.S
Hence \sin^{-1}\frac{63}{65}= \sin^{-1}\frac{5}{13}+\cos^{-1}\frac{3}{5}
 

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