# then x is equal to:

Question:

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

1. (1) $\frac{\sqrt{145}}{12}$

2. (2) $\frac{\sqrt{145}}{10}$

3. (3) $\frac{\sqrt{146}}{12}$

4. (4) $\frac{\sqrt{145}}{11}$

Correct Option: 1

Solution:

$\cos ^{-1}\left(\frac{2}{3 x}\right)+\cos ^{-1}\left(\frac{3}{4 x}\right)=\frac{\pi}{2} ;\left(x>\frac{3}{4}\right)$

$\Rightarrow \quad \cos ^{-1}\left(\frac{2}{3 x}\right)=\frac{\pi}{2}-\cos ^{-1}\left(\frac{3}{4 x}\right)$

$\Rightarrow \quad \cos ^{-1}\left(\frac{2}{3 x}\right)=\sin ^{-1}\left(\frac{3}{4 x}\right)$

$\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$

Put $\sin ^{-1}\left(\frac{3}{4 x}\right)=\theta \quad \Rightarrow \sin \theta=\frac{3}{4 x}$

$\Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta}=\sqrt{1-\frac{9}{16 x^{2}}}$

$\Rightarrow \theta=\cos ^{-1}\left(\frac{\sqrt{16 x^{2}-9}}{4 x}\right)$

$\therefore \quad \cos ^{-1}\left(\frac{2}{3 x}\right)=\cos ^{-1}\left(\frac{\sqrt{16 x^{2}-9}}{4 x}\right)$

$\Rightarrow \quad \frac{2}{3 x}=\frac{\sqrt{16 x^{2}-9}}{4 x} \Rightarrow x^{2}=\frac{64+81}{9 \times 16} \Rightarrow x=\pm \sqrt{\frac{145}{144}}$

$\Rightarrow \quad x=\frac{\sqrt{145}}{12}$ $\left(\because x>\frac{3}{4}\right)$