Section - C Q. Nos. 27 to 32 carry 4 marks each. Solve the equation x: \sin ^{-1} (\frac{5}{x}) + \sin ^{-1} (\frac{12}{x}) =\frac{\pi}{2}(x \neq 0)

CBSE Class 12 Math 2020 Delhi Set 3 Solved Paper - Question 9

Question : 9 of 11

Marks: +1, -0

Section - C

Q. Nos. 27 to 32 carry 4 marks each.
Solve the equation

x: sin \;^{-1} (\;{5}/{x}) + sin \;^{-1} (\;{12}/{x}) =\;{π}/{2}(x ≠ 0)

Solution:

sin \;^{-1} \;{5}/{x}+ sin \;^{-1} \;{12}/{x}=\;{π}/{2}

⇒ \;{12}/{x}= sin \; (\;{π}/{2}- sin \;^{-1} \;{5}/{x})

⇒ \;{12}/{x}=\cos ( sin \;^{-1} \;{5}/{x})

⇒ \;{12}/{x}=√{1- (\;{5}/{x}) ^2} ∵ \cos ( sin \;^{-1} x) ={√{1-x^2}}

(\;{12}/{x}) ^2=1-\;{25}/{x^2}

⇒ \;{144}/{x^2}+\;{25}/{x^2}=1

⇒ \;{169}/{x^2}=1

x^2=169

⇒ {\x}= ± {√{169}}

⇒ {\x}=13,-13

But

{\x}=-13

doesn't satisfy the given equation.
Hence the solution of the given equation is

{\x}=13

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