Animated Solution for Mathematics - Inverse Trigonometric Functions: If ${\cos }^{-1}(\frac{2}{3x})+{\cos }^{-1}(\frac{3}{4x})=\frac{\pi }{2}(x>\frac{3}{4})$, then x is equal to :

Welcome, fellow traveler of the JEE path. Today, we are not just solving an equation; we are uncovering a hidden symmetry.

When you look at the expression ${\cos }^{-1}(\frac{2}{3x})+{\cos }^{-1}(\frac{3}{4x})=\frac{\pi }{2}$, do not see it as a wall of symbols. See it as a balance scale.

We have two angles, let us call them $\alpha$ and $\beta$, whose sum is exactly $\frac{\pi }{2}$. In the world of geometry, this is a profound statement. It tells us that these two angles are complementary, fitting together perfectly to form a right angle.

The Power of Transformation

Our first instinct might be to panic at the sight of inverse functions. But remember, an inverse function is just a fancy way of saying 'the angle whose...'.

Let us isolate one term to make our lives easier. By moving ${\cos }^{-1}(\frac{3}{4x})$ to the right side, we get:

Now, look at the right side. Does it ring a bell? We know the fundamental identity ${\sin }^{-1}(\theta )+{\cos }^{-1}(\theta )=\frac{\pi }{2}$.

This is the bridge between the two worlds of sine and cosine. By applying this, our equation transforms into:

Suddenly, the complexity evaporates. We are no longer dealing with a sum; we are dealing with an equivalence.

The Geometry of the Triangle

Let us define $\theta ={\sin }^{-1}(\frac{3}{4x})$. This means $\sin (\theta )=\frac{3}{4x}$.

Imagine a right-angled triangle where the side opposite to $\theta$ is $3$ and the hypotenuse is $4x$. To find the cosine of this same angle, we need the adjacent side. Using the Pythagorean theorem, we find the base:

Therefore, $\cos (\theta )=\frac{\sqrt{16x^2-9}}{4x}$. Now, we can rewrite our equation as:

The Algebraic Resolution

Since the cosine function is one-to-one within its principal domain, we can simply equate the arguments:

Here is where the magic happens. Because we are given $x>\frac{3}{4}$, we know $x$ is not zero. We can safely cancel the $x$ from the denominators, leaving us with a clean, manageable equation:

Cross-multiplying gives us $\frac{8}{3}=\sqrt{16x^2-9}$. Now, we square both sides to set our variable free:

Adding $9$ to both sides (which is $\frac{81}{9}$), we get:

Finally, dividing by $16$ and taking the square root, we arrive at our destination:

Look at what we have done. We started with a daunting inverse trigonometric equation and, through the simple application of complementary identities and the Pythagorean theorem, reduced it to basic arithmetic.

This is the essence of JEE mathematics: identifying the underlying structure, applying the right tool, and trusting the process. You have mastered the symmetry. Keep this confidence, and carry it into your next challenge! The final answer is $x=\frac{\sqrt{145}}{12}$.