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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.12 Question 1 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.12 Question 1 Maths Textbook Solution.

Answers (1)

Answer:\frac{33}{65}        
Given:
             \cos \left ( \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13} \right )
Hint: 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 ]
Solution:  Using the formula

\sin^{-1}x+\sin^{-1}y= \sin^{-1}\left [ x\sqrt{1-y^{2}}+y\sqrt{1-x^{2}} \right ]
Substituting the value we get,
\begin{aligned} &\cos ^{-1}\left[\sin ^{-1}\left(\frac{3}{5} \sqrt{1-\left(\frac{5}{13}\right)^{2}}+\frac{5}{13} \sqrt{1-\left(\frac{3}{5}\right)^{2}}\right)\right] \\ &=\cos ^{-1}\left[\sin ^{-1}\left(\frac{3}{5} \times \frac{12}{13}+\frac{5}{13} \times \frac{4}{5}\right)\right] \\ &=\cos ^{-1}\left[\sin ^{-1}\left(\frac{36}{65}+\frac{20}{65}\right)\right] \\ &=\cos ^{-1}\left[\sin ^{-1}\left(\frac{56}{65}\right)\right] \end{aligned}
Again, we know that
\sin^{-1}x= \cos^{-1}\sqrt{1-x^{2}}
Now, substituting we get
\begin{aligned} &=\cos \left[\cos ^{-1} \sqrt{1-\left(\frac{56}{65}\right)^{2}}\right] \\ &=\cos \left[\cos ^{-1} \sqrt{\left(\frac{33}{65}\right)}\right] \end{aligned}
= \frac{33}{65}\; \; \; \; \; \; \; \; \; \left [ \because \cos \left ( \cos^{-1} \right )= x \right ]                                                                                    
Hence        \cos \left ( \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{5}{13} \right )= \frac{33}{65}

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