Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of $\tan \left[{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{2}{3}\right)\right]$ is

The problem asks us to evaluate the expression:

At first glance, this looks like a jumble of inverse functions. However, remember that an inverse trigonometric function is simply an angle.

Let us define the first angle as $\theta ={\cos }^{-1}\left(\frac{4}{5}\right)$. This implies that $\cos \theta =\frac{4}{5}$.

In a right-angled triangle, $\cos \theta$ is defined as the ratio of the base to the hypotenuse. If we place $\theta$ at the base, our adjacent side is $4$ and our hypotenuse is $5$.

Using the Pythagorean theorem, the perpendicular side is:

Now, we can express our angle $\theta$ in terms of tangent:

Therefore, we have successfully translated the language of cosine into the language of tangent:

Now that our expression is $\tan \left[{\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{2}{3}\right)\right]$, the path forward becomes clear. We are dealing with the sum of two inverse tangents.

The JEE curriculum provides us with a powerful tool for this: the addition identity:

Here, our $A=\frac{3}{4}$ and our $B=\frac{2}{3}$. Let us carefully substitute these into our identity.

The numerator becomes:

The denominator becomes:

We are now left with the following expression:

Notice how the denominators of $12$ cancel out beautifully, leaving us with:

By the fundamental property of inverse functions, $\tan ({\tan }^{-1}x)=x$. Thus, our final result is:

In the world of competitive exams, sometimes the answer is simply 'none' if the options provided do not match. Trust your derivation, trust your math, and embrace the result!