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

Solving Inverse Trigonometry

Welcome, future engineer. Today, we are going to dismantle a problem that often trips up students in the JEE Advanced examination. It looks intimidating—a $\cot$ function wrapping around a sum of two different inverse trigonometric functions.

But here is the secret: trigonometry is not about memorizing formulas; it is about geometry and unification.

The Triangle Strategy

Our first task is to make the expression speak one language. We have ${\text{cosec}}^{-1}\frac{5}{3}$ and ${\tan }^{-1}\frac{2}{3}$. We cannot add these directly.

Let us focus on $\theta ={\text{cosec}}^{-1}\frac{5}{3}$. By definition, this means $\text{cosec}\theta =\frac{5}{3}$.

Imagine a right-angled triangle where the hypotenuse is $5$ and the perpendicular side is $3$. Using the Pythagoras theorem, ${\text{Base}}^2={\text{Hypotenuse}}^2-{\text{Perpendicular}}^2$, we find:

Thus, the base is $4$. Now, we can express $\theta$ in terms of tangent:

Therefore, $\theta ={\tan }^{-1}\frac{3}{4}$. We have successfully translated the first term!

The Addition Engine

Now our expression looks much cleaner: $\cot \left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{2}{3}\right)$. We are now dealing with the sum of two ${\tan }^{-1}$ functions.

We invoke the powerful identity:

Here, $x=\frac{3}{4}$ and $y=\frac{2}{3}$. Before we proceed, we check the condition $xy=\frac{3}{4}\times \frac{2}{3}=\frac{1}{2}<1$. The condition holds, so we proceed with confidence.

Substituting the values, we get the argument inside the ${\tan }^{-1}$ as:

The Elegant Collapse

Let us simplify this fraction with care. The numerator is:

The denominator is:

When we divide these, the $12$ in the denominators cancels out perfectly, leaving us with $\frac{17}{6}$. Our expression has now collapsed to $\cot \left({\tan }^{-1}\frac{17}{6}\right)$.

To finish, we use the reciprocal property: ${\tan }^{-1}\frac{17}{6}={\cot }^{-1}\frac{6}{17}$. Finally:

We have arrived at the answer. Notice how the complexity vanished once we chose the right path. Keep this systematic approach in your toolkit, and no trigonometry problem will ever stand in your way.