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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.12 Question 3 Subquestion (iii) Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.12 Question 3 Subquestion (iii) Maths Textbook Solution.

Answers (1)

Answer:  +13
Given: \sin^{-1}\frac{5}{x}+\sin^{-1}\frac{12}{x}= \frac{\pi }{2}
Hint: Here we will use \frac{\pi }{2}-\sin^{-1}x= \cos^{-1}x
Solution: Here we have
\sin^{-1}\frac{5}{x}+\sin^{-1}\frac{12}{x}= \frac{\pi }{2}
\sin^{-1}\frac{5}{x}= \frac{\pi }{2}-\sin^{-1}\frac{12}{x}
\Rightarrow \sin^{-1}\frac{5}{x}= \cos^{-1}\frac{12}{x} \; \; \; \; \left [ \because \frac{\pi }{2}-\sin^{-1}x= \cos^{-1}x \right ]                                                    
\Rightarrow \sin^{-1}\frac{5}{x}=\sin^{-1}\left ( \sqrt{1-\left ( \frac{12}{x} \right )^{2}} \right ) \; \; \; \; \; \; \; \; \; \; [\cos^{-1}x=\sin^{-1}\sqrt{1-x^{2}}]
\Rightarrow \left ( \frac{5}{x}\right )^{2} =\sqrt{1-\left ( \frac{12}{x} \right )^{2}}
Squaring on both sides, we get
 \Rightarrow \left ( \frac{5}{x}\right )^{2} =1-\left ( \frac{12}{x} \right )^{2}
\Rightarrow \left ( \frac{25}{x^{2}} \right )+\left ( \frac{144}{x^{2}} \right )= 1
\Rightarrow x^{2}= 169
\Rightarrow x=\pm 13 [ As -13 is not satisfying the equation]
Hence x= 13 is the result

 

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