Question

# If $$\cos^{-1} \left (\dfrac {2}{3x}\right ) + \cos^{-1} \left (\dfrac {3}{4x}\right ) = \dfrac {\pi}{2}\left (x > \dfrac {3}{4}\right )$$ then $$x$$ is equal to

A
14512
B
14510
C
14612
D
14511

Solution

## The correct option is A $$\dfrac {\sqrt {145}}{12}$$$$\cos^{-1}\left (\dfrac {2}{3x}\right ) + \cos^{-1} \left (\dfrac {3}{4x}\right ) = \dfrac {\pi}{2}\left (x > \dfrac {3}{4}\right )$$$$\Rightarrow \cos^{-1} \left (\dfrac {3}{4x}\right ) = \dfrac {\pi}{2} - \cos^{-1} \left (\dfrac {2}{3x}\right )$$$$\Rightarrow \cos^{-1} \left (\dfrac {3}{4x}\right ) = \sin^{-1} \left (\dfrac {2}{3x}\right )$$$$\Rightarrow \cos \left (\cos^{-1} \left (\dfrac {3}{4x}\right )\right ) = \cos \left (\sin^{-1}\dfrac {2}{3x}\right )$$$$\Rightarrow\dfrac {3}{4x} = \dfrac {\sqrt {9x^{2} - 4}}{3x}$$$$\Rightarrow\dfrac {81}{16} + 4 = 9x^{2}$$$$\Rightarrow x^{2} = \dfrac {145}{16 \times 9}\Rightarrow x = \dfrac {\sqrt {145}}{12}$$.Mathematics

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