Inverse Circular Functions 2 Question 3
3. If $\cos ^{-1} \frac{2}{3 x}+\cos ^{-1} \frac{3}{4 x}=\frac{\pi}{2} x>\frac{3}{4}$, then $x$ is equal to
(a) $\frac{\sqrt{145}}{10}$
(b) $\frac{\sqrt{146}}{12}$
(c) $\frac{\sqrt{145}}{12}$
(d) $\frac{\sqrt{145}}{11}$
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Answer:
Correct Answer: 3. (c)
Solution:
Key Idea Use the formula,
$\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)$
We have, $\quad \cos ^{-1} \frac{2}{3 x}+\cos ^{-1} \frac{3}{4 x}=\frac{\pi}{2}$
$\Rightarrow \cos ^{-1} \frac{2}{3 x} \cdot \frac{3}{4 x}-\sqrt{1-\frac{4}{9 x^{2}}} \sqrt{1-\frac{9}{16 x^{2}}}=\frac{\pi}{2}$
$\left[\because \cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)\right]$
$\Rightarrow \quad \cos ^{-1} \frac{1}{2 x^{2}}-\frac{\sqrt{9 x^{2}-4} \sqrt{16 x^{2}-9}}{12 x^{2}}=\frac{\pi}{2}$
$\Rightarrow \quad \frac{6-\sqrt{9 x^{2}-4} \sqrt{16 x^{2}-9}}{12 x^{2}}=\cos \frac{\pi}{2}=0$
$\Rightarrow \sqrt{9 x^{2}-4} \sqrt{16 x^{2}-9}=6$
On squaring both sides,
$\Rightarrow \quad\left(9 x^{2}-4\right)\left(16 x^{2}-9\right)=36$
$\Rightarrow \quad 144 x^{4}-81 x^{2}-64 x^{2}+36=36$
$\Rightarrow \quad 144 x^{4}-145 x^{2}=0$
$\Rightarrow \quad x^{2}\left(144 x^{2}-145\right)=0$
$\Rightarrow \quad x=0$ or $x= \pm \sqrt{\frac{145}{144}}= \pm \frac{\sqrt{145}}{12}$
But $x>\frac{3}{4}$,
$$ x=\frac{\sqrt{145}}{12} $$
