JEE PYQ: Inverse Trigonometric Functions Question 17
Question 17 - 2019 (09 Jan Shift 1)
If $\cos^{-1}\left(\frac{2}{3x}\right) + \cos^{-1}\left(\frac{3}{4x}\right) = \frac{\pi}{2}$ $\left(x > \frac{3}{4}\right)$, then $x$ is equal to:
(1) $\frac{\sqrt{145}}{12}$
(2) $\frac{\sqrt{145}}{10}$
(3) $\frac{\sqrt{146}}{12}$
(4) $\frac{\sqrt{145}}{11}$
Show Answer
Answer: (1)
Solution
$\cos^{-1}\frac{2}{3x} = \frac{\pi}{2} - \cos^{-1}\frac{3}{4x} = \sin^{-1}\frac{3}{4x}$. Put $\sin^{-1}\frac{3}{4x} = \theta$, so $\sin\theta = \frac{3}{4x}$ and $\cos\theta = \sqrt{1 - \frac{9}{16x^2}}$. Also $\cos^{-1}\frac{2}{3x} = \theta$ means $\cos\theta = \frac{2}{3x}$. So $\frac{2}{3x} = \frac{\sqrt{16x^2-9}}{4x}$. Squaring: $x^2 = \frac{64+81}{9\times 16} = \frac{145}{144}$. Hence $x = \frac{\sqrt{145}}{12}$.
BETA
