if cos−1⁡23x+cos−1⁡34x=π2x>34 then x is equal to

A
$\frac{\sqrt{145}}{12}$
B
$\frac{\sqrt{145}}{11}$
C
$\frac{\sqrt{145}}{10}$
D
$\frac{\sqrt{146}}{12}$

Solution:

We have, $\cos^{- 1} (\frac{2}{3 x}) + \cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2}$

$\begin{array}{l} \Rightarrow \cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2} - \cos^{- 1} (\frac{2}{3 x}) \\ \Rightarrow \cos^{- 1} (\frac{3}{4 x}) = \sin^{- 1} (\frac{2}{3 x}) \end{array}$

$\begin{array}{l} \Rightarrow \cos (\cos^{- 1} (\frac{3}{4 x})) = \cos (\sin^{- 1} (\frac{2}{3 x})) \\ \Rightarrow \frac{3}{4 x} = \sqrt{1 - \frac{4}{(3 x)^{2}}} \Rightarrow \frac{3}{4} = \frac{\sqrt{9 x^{2} - 4}}{3} \end{array}$

Squaring both sides, we get

$\frac{9}{16} = \frac{9 x^{2} - 4}{9} \Rightarrow x^{2} = \frac{145}{144} \Rightarrow x = \frac{\sqrt{145}}{12}$

if $\cos^{- 1} (\frac{2}{3 x}) + \cos^{- 1} (\frac{3}{4 x}) = \frac{π}{2} (x > \frac{3}{4})$ then $x$ is equal to

Solution:

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