Question

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

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

Answer 1

Answer: (A)

$\frac{\sqrt{145}}{12}$

Answer 2

The given equation is $\cos^{-1}(\frac{2}{3x})+\cos^{-1}(\frac{3}{4x})=\frac{\pi}{2}$. We are given that $x > \frac{3}{4}$.

For the terms $\cos^{-1}(\frac{2}{3x})$ and $\cos^{-1}(\frac{3}{4x})$ to be defined, their arguments must be in the interval $[-1, 1]$.

We can use the property of inverse trigonometric functions: $\cos^{-1} A + \cos^{-1} B = \frac{\pi}{2}$ if and only if $A^2 + B^2 = 1$, provided $A \ge 0$, $B \ge 0$.

Here, $A = \frac{2}{3x}$ and $B = \frac{3}{4x}$. Since $x > \frac{3}{4}$, both $A$ and $B$ are positive. The condition $A^2+B^2=1$ must hold for the sum to be $\frac{\pi}{2}$.

$(\frac{2}{3x})^2 + (\frac{3}{4x})^2 = 1$

$\frac{4}{9x^2} + \frac{9}{16x^2} = 1$

To combine the terms on the left side, find a common denominator, which is $144x^2$:

$\frac{4 \times 16}{9 \times 16 x^2} + \frac{9 \times 9}{16 \times 9 x^2} = 1$

$\frac{64}{144x^2} + \frac{81}{144x^2} = 1$

$\frac{64 + 81}{144x^2} = 1$

$\frac{145}{144x^2} = 1$

$145 = 144x^2$

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

Taking the square root of both sides:

$x = \pm \sqrt{\frac{145}{144}} = \pm \frac{\sqrt{145}}{12}$.

We are given the condition $x > \frac{3}{4}$. We need to check which of the two possible values of $x$ satisfies this condition. The positive value is $x = \frac{\sqrt{145}}{12}$.

To check if $\frac{\sqrt{145}}{12} > \frac{3}{4}$, we can compare their squares:

$(\frac{\sqrt{145}}{12})^2 = \frac{145}{144}$

$(\frac{3}{4})^2 = \frac{9}{16} = \frac{9 \times 9}{16 \times 9} = \frac{81}{144}$.

Since $\frac{145}{144} > \frac{81}{144}$, we have $(\frac{\sqrt{145}}{12})^2 > (\frac{3}{4})^2$.

Since both $\frac{\sqrt{145}}{12}$ and $\frac{3}{4}$ are positive, taking the square root preserves the inequality:

$\frac{\sqrt{145}}{12} > \frac{3}{4}$.

So, the value $x = \frac{\sqrt{145}}{12}$ satisfies the given condition. The negative value $x = -\frac{\sqrt{145}}{12}$ is clearly not greater than $\frac{3}{4}$.

Thus, the only valid solution is $x = \frac{\sqrt{145}}{12}$.