Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

We have,${\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}\frac{2}{3x}+{\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}\frac{3}{4x}=\frac{\pi }{2}, x>\frac{3}{4} ...\left(\mathrm{i}\right)$

$\therefore {\cos }^{-1}\frac{3}{4x}={\sin }^{-1}\frac{\sqrt{16x^2-9}}{4x}$ 

Put in equation $\left(\mathrm{i}\right)$ we get, 

$\Rightarrow {\mathrm{c}\mathrm{o}\mathrm{s}}^{-1}\frac{2}{3x}+{\mathrm{s}\mathrm{i}\mathrm{n}}^{-1}\frac{\sqrt{16x^2-9}}{4x}=\frac{\pi }{2} ...\left(\mathrm{ii}\right)$ 

We know ${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

From equation $\left(\mathrm{ii}\right)$ we get, 

$\Rightarrow \frac{2}{3x}=\frac{\sqrt{16x^2-9}}{4x}$

$\Rightarrow {\left(\frac{8}{3}\right)}^2=16x^2-9 \left(\because x\ne 0\right)$

$\Rightarrow 16x^2=\frac{64}{9}+9=\frac{145}{9}$

$\Rightarrow x^2=\frac{145}{144}$

$\Rightarrow x=\frac{\sqrt{145}}{12}$