Animated Solution for Mathematics - Inverse Trigonometric Functions: If we consider only the principle values of the inverse trigonometric functions then the value of $\tan \left({\cos }^{-1}\frac{1}{5\sqrt{2}}-{\sin }^{-1}\frac{4}{\sqrt{17}}\right)$ is

Welcome, future engineer. Today, we aren't just solving a problem; we are peeling back the layers of inverse trigonometry to reveal the simple geometric truth hidden underneath.

When you see an expression like $\tan \left({\cos }^{-1}\frac{1}{5\sqrt{2}}-{\sin }^{-1}\frac{4}{\sqrt{17}}\right)$, it is natural to feel a bit intimidated. But remember, inverse trigonometric functions are just angles.

Let us define $\alpha ={\cos }^{-1}\frac{1}{5\sqrt{2}}$ and $\beta ={\sin }^{-1}\frac{4}{\sqrt{17}}$. Our goal is simply to find $\tan (\alpha -\beta )$.

To find the tangent of these angles, we don't need complex calculus; we need the ancient, reliable power of right-angled triangles.

For $\alpha$, we know $\cos \alpha =\frac{1}{5\sqrt{2}}$. In a right triangle, cosine is $\frac{\text{Base}}{\text{Hypotenuse}}$. So, let the base be $1$ and the hypotenuse be $5\sqrt{2}$.

Using Pythagoras' theorem, the perpendicular side is:

Thus, $\tan \alpha =\frac{7}{1}=7$.

Similarly, for $\beta$, we have $\sin \beta =\frac{4}{\sqrt{17}}$. Sine is $\frac{\text{Perpendicular}}{\text{Hypotenuse}}$. Let the perpendicular be $4$ and the hypotenuse be $\sqrt{17}$.

The base is:

Thus, $\tan \beta =\frac{4}{1}=4$.

Now, the problem has transformed into something beautiful. We have $\alpha ={\tan }^{-1}(7)$ and $\beta ={\tan }^{-1}(4)$. We need to calculate $\tan (\alpha -\beta )$.

We reach into our mathematical toolkit and pull out the tangent subtraction identity:

Substituting our values, we get:

The numerator is $3$, and the denominator is $1+28=29$.

The final result is $\frac{3}{29}$. It is a perfect, clean fraction.

You see, the complexity was just a mask. By breaking it down into triangles and applying the right identity, we turned a daunting expression into a simple arithmetic problem. Keep this mindset, and no JEE problem will ever be too difficult for you.