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CBSE Class 12 Math 2020 Delhi Set 3 Solved Paper

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Question : 9 of 11
Marks: +1, -0
Section - C
Q. Nos. 27 to 32 carry 4 marks each.
Solve the equation x:sin⁡−1(5x)+sin⁡−1(12x)=π2(x≠0)x: \sin^{-1} \left( \frac{5}{x} \right) + \sin^{-1} \left( \frac{12}{x} \right) = \frac{\pi}{2} \quad (x \neq 0)
Solution:  
sin⁡−15x+sin⁡−112x=π2\sin^{-1} \frac{5}{x} + \sin^{-1} \frac{12}{x} = \frac{\pi}{2}
⇒12x=sin⁡(π2−sin⁡−15x)\Rightarrow \frac{12}{x} = \sin \left( \frac{\pi}{2} - \sin^{-1} \frac{5}{x} \right)
⇒12x=cos⁡(sin⁡−15x)\Rightarrow \frac{12}{x} = \cos \left( \sin^{-1} \frac{5}{x} \right)
⇒12x=1−(5x)2∵cos⁡(sin⁡−1x)=1−x2\Rightarrow \frac{12}{x} = \sqrt{1 - \left( \frac{5}{x} \right)^2} \because \cos(\sin^{-1} x) = \sqrt{1-x^2}
(12x)2=1−25x2\left( \frac{12}{x} \right)^2 = 1 - \frac{25}{x^2}
⇒144x2+25x2=1\Rightarrow \frac{144}{x^2} + \frac{25}{x^2} = 1
⇒169x2=1\Rightarrow \frac{169}{x^2} = 1
x2=169x^2 = 169
⇒x=±169\Rightarrow x = \pm \sqrt{169}
⇒x=13,−13\Rightarrow x = 13, -13
But x=−13x = -13 doesn't satisfy the given equation.
Hence the solution of the given equation is x=13x = 13.
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