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Question: If \({{\sin }^{-1}}\left( \dfrac{5}{x} \right)+{{\sin }^{-1}}\left( \dfrac{12}{x} \right)=\dfrac{\pi...

If sin1(5x)+sin1(12x)=π2{{\sin }^{-1}}\left( \dfrac{5}{x} \right)+{{\sin }^{-1}}\left( \dfrac{12}{x} \right)=\dfrac{\pi }{2}, the find the value of xx.
A. 713\dfrac{7}{13}
B. 43\dfrac{4}{3}
C. 13
D. 137\dfrac{13}{7}

Explanation

Solution

Hint : We first use the associative angle formula to find the simplified from. Then we use the trigonometric ratio of the right-angle triangle to find the value of xx.

Complete step-by-step answer :
We simplify the equation by taking one ratio of sin on the other side
sin1(5x)+sin1(12x)=π2{{\sin }^{-1}}\left( \dfrac{5}{x} \right)+{{\sin }^{-1}}\left( \dfrac{12}{x} \right)=\dfrac{\pi }{2}.
So,
sin1(12x)=π2sin1(5x){{\sin }^{-1}}\left( \dfrac{12}{x} \right)=\dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{5}{x} \right).
Now we take ratio of sine on both sides to get
sin[sin1(12x)]=sin[π2sin1(5x)]\sin \left[ {{\sin }^{-1}}\left( \dfrac{12}{x} \right) \right]=\sin \left[ \dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{5}{x} \right) \right].
We get
sin[π2sin1(5x)]=12x=cos[sin1(5x)]\sin \left[ \dfrac{\pi }{2}-{{\sin }^{-1}}\left( \dfrac{5}{x} \right) \right]=\dfrac{12}{x}=\cos \left[ {{\sin }^{-1}}\left( \dfrac{5}{x} \right) \right].
We can take the representation of a right-angle triangle with height and hypotenuse ratio being (5x)\left( \dfrac{5}{x} \right) and the angle being θ\theta . The height and base were considered with respect to that particular angle θ\theta .

In this case we take BC=xBC=x and keeping the ratio in mind we have AC=5AC=5 as the ratio has to be (5x)\left( \dfrac{5}{x} \right).
Now we apply the Pythagoras’ theorem to find the length of AB. BC2=AB2+AC2B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}.
So, AB2=52x2A{{B}^{2}}={{5}^{2}}-{{x}^{2}} which gives AB=x225AB=\sqrt{{{x}^{2}}-25}.
We need to find cos[sin1(5x)]\cos \left[ {{\sin }^{-1}}\left( \dfrac{5}{x} \right) \right] which is equal to cosθ\cos \theta .
This ratio gives cosθ=basehypotenuse\cos \theta =\dfrac{\text{base}}{\text{hypotenuse}}. So,
cosθ=ABBC=x225x\cos \theta =\dfrac{AB}{BC}=\dfrac{\sqrt{{{x}^{2}}-25}}{x}.
cos[sin1(5x)]=12x x225x=12x \begin{aligned} & \cos \left[ {{\sin }^{-1}}\left( \dfrac{5}{x} \right) \right]=\dfrac{12}{x} \\\ & \Rightarrow \dfrac{\sqrt{{{x}^{2}}-25}}{x}=\dfrac{12}{x} \\\ \end{aligned}
Simplification gives x225=122=144x=13{{x}^{2}}-25={{12}^{2}}=144\Rightarrow x=13. The correct option is C.
So, the correct answer is “Option C”.

Note : We can also apply the trigonometric image form to get the value of cos[sin1(5x)]\cos \left[ {{\sin }^{-1}}\left( \dfrac{5}{x} \right) \right].
It’s given that sinθ=5x\sin \theta =\dfrac{5}{x} and we need to find cosθ\cos \theta . We know cosθ=1sin2θ\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }.
Putting the values, we get cosθ=1sin2θ=1(5x)2=x225x\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-{{\left( \dfrac{5}{x} \right)}^{2}}}=\dfrac{\sqrt{{{x}^{2}}-25}}{x}.