Animated Solution for Mathematics - Inverse Trigonometric Functions: The value of ${\tan }^{-1}\left(\frac{\cos \left(\frac{15\pi }{4}\right)-1}{\sin \left(\frac{\pi }{4}\right)}\right)$ is equal to

Welcome, fellow traveler on the JEE journey. Today, we confront a problem that often makes students pause: the evaluation of:

At first glance, that angle $\frac{15\pi }{4}$ feels like a mountain. But in mathematics, as in life, the biggest obstacles are often just illusions of scale. Let us break this down.

The angle $\frac{15\pi }{4}$ is simply a point on the unit circle. To understand it, we use the periodicity of the cosine function, knowing that $\cos (\theta )=\cos (2n\pi +\theta )$.

We can rewrite $\frac{15\pi }{4}$ as $4\pi -\frac{\pi }{4}$. Since $4\pi$ represents two full revolutions, we are effectively at the same position as $-\frac{\pi }{4}$.

Because cosine is an even function, $\cos (-\frac{\pi }{4})=\cos (\frac{\pi }{4})$. We have successfully reduced a complex angle to the familiar $\frac{\pi }{4}$, which has a value of $\frac{1}{\sqrt{2}}$.

Now, look at our expression:

This is where many students stumble, not because of the trigonometry, but because of the algebra. Multiply both the numerator and the denominator by $\sqrt{2}$.

The numerator becomes $1-\sqrt{2}$, and the denominator becomes $1$. Our expression is now simply ${\tan }^{-1}(1-\sqrt{2})$.

This is the moment of truth. We need to find an angle $\theta$ such that $\tan (\theta )=1-\sqrt{2}$.

You might recall that $\tan (\frac{\pi }{8})=\sqrt{2}-1$. Our value is the negative of this.

Using the odd property of the tangent function, $\tan (-\theta )=-\tan (\theta )$, we see that:

We are left with ${\tan }^{-1}(\tan (-\frac{\pi }{8}))$. Since $-\frac{\pi }{8}$ lies within the principal value branch of the inverse tangent function, which is $(-\frac{\pi }{2},\frac{\pi }{2})$, we can safely conclude the result.

The final answer is $-\frac{\pi }{8}$.

You have navigated the complexity, simplified the expression, and arrived at the elegant truth. Keep this confidence with you; every problem is just a series of small, manageable steps.