Animated Solution for Mathematics - Inverse Trigonometric Functions: $\tan \left(2{\tan }^{-1}\frac{1}{5}+{\sec }^{-1}\frac{\sqrt{5}}{2}+2{\tan }^{-1}\frac{1}{8}\right)$ is equal to:

Welcome, future engineer! Today, we are going to dismantle a trigonometric expression that looks like a tangled mess of inverse functions.

At first glance, the expression might seem like a nightmare, but I want you to see it as a puzzle waiting to be solved. The secret to mastering these problems is not brute force, but strategic grouping and the use of a 'universal language.'

Look closely at the expression. We have two terms with a coefficient of $2$. In the world of JEE mathematics, whenever you see a common coefficient, your brain should immediately scream 'Group them!'

By factoring out the $2$, we transform the expression into:

Now, we use the addition identity to simplify the part inside the parentheses. Plugging in $x=\frac{1}{5}$ and $y=\frac{1}{8}$, we get:

Simplifying the numerator gives us $\frac{13}{40}$, and the denominator becomes $1-\frac{1}{40}=\frac{39}{40}$. The $40$ cancels out, leaving us with ${\tan }^{-1}\left(\frac{13}{39}\right)$, which is simply ${\tan }^{-1}\frac{1}{3}$.

Now, we are left with $\tan \left[2{\tan }^{-1}\frac{1}{3}+{\sec }^{-1}\frac{\sqrt{5}}{2}\right]$. We have a ${\sec }^{-1}$ term that doesn't fit in with our ${\tan }^{-1}$ family, so we must convert it.

Let $\theta ={\sec }^{-1}\frac{\sqrt{5}}{2}$, which implies $\sec \theta =\frac{\sqrt{5}}{2}$. Imagine a right-angled triangle where the hypotenuse is $\sqrt{5}$ and the base is $2$.

By the Pythagorean theorem, the perpendicular is $\sqrt{(\sqrt{5}{)}^2-2^2}=\sqrt{5-4}=1$. Thus, $\tan \theta =\frac{\text{perpendicular}}{\text{base}}=\frac{1}{2}$. Our expression is now:

We use the double-angle identity $2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$. With $x=\frac{1}{3}$, this becomes:

The expression has finally collapsed into $\tan \left({\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{1}{2}\right)$. One last application of the addition identity yields:

The numerator is $\frac{3+2}{4}=\frac{5}{4}$, and the denominator is $1-\frac{3}{8}=\frac{5}{8}$. Dividing $\frac{5}{4}$ by $\frac{5}{8}$ gives us $2$.

The final expression is $\tan ({\tan }^{-1}2)$, which is simply $2$. You have successfully navigated the complexity and arrived at the elegant solution.