Animated Solution for Mathematics - Inverse Trigonometric Functions: $50\tan \left(3{\tan }^{-1}\left(\frac{1}{2}\right)+2{\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)+4\sqrt{2}\tan \left(\frac{1}{2}{\tan }^{-1}(2\sqrt{2})\right)$ is equal to \_\_\_\_.

The given expression is:

We divide this into two manageable parts: $T_1$ and $T_2$.

Taming the First Beast

Let $\alpha ={\tan }^{-1}(1/2)$ and $\beta ={\cos }^{-1}(1/\sqrt{5})$. By definition, $\tan \alpha =1/2$.

For $\beta$, we visualize a right-angled triangle where the base is $1$ and the hypotenuse is $\sqrt{5}$. Using the Pythagorean theorem, the perpendicular is $\sqrt{(\sqrt{5}{)}^2-1^2}=2$. Thus, $\tan \beta =2/1=2$.

Observe the relationship: $\tan \alpha \cdot \tan \beta =(1/2)\cdot 2=1$. This implies that $\alpha +\beta =\pi /2$, or $\beta =\pi /2-\alpha$.

Substituting this into the argument of the first term:

The expression simplifies to $50\tan (\pi +\alpha )$. Since $\tan (\pi +\theta )=\tan \theta$, this becomes $50\tan \alpha$.

Substituting $\tan \alpha =1/2$, we find:

The Half-Angle Challenge

Now, consider $T_2=4\sqrt{2}\tan \left(\frac{1}{2}{\tan }^{-1}(2\sqrt{2})\right)$. Let $\theta ={\tan }^{-1}(2\sqrt{2})$, so $\tan \theta =2\sqrt{2}$.

We need to find $t=\tan (\theta /2)$. We invoke the half-angle identity:

Substituting our value, we get $2\sqrt{2}=\frac{2t}{1-t^2}$, which simplifies to the quadratic equation:

Solving this using the quadratic formula:

This yields two potential values: $t=1/\sqrt{2}$ or $t=-\sqrt{2}$. Since $\theta$ is an acute angle, $\theta /2$ must also be acute, so we reject the negative root.

Thus, $t=1/\sqrt{2}$, and the second term becomes:

We have conquered both parts of the expression. The total value is:

It is a testament to the beauty of mathematics that such an intimidating expression collapses into a simple integer. Remember, in the JEE Advanced exam, never let the complexity of the notation blind you to the underlying geometric simplicity.