Animated Solution for Mathematics - Limits, Continuity and Differentiability: The value of the limit ${\lim }_{\theta \to 0}\frac{\tan (\pi {\cos }^2\theta )}{\sin (2\pi {\sin }^2\theta )}$ is equal to :

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dissect a limit problem that, at first glance, might seem like a chaotic mess of trigonometric functions.

But as we peel back the layers, you will see that it is actually a beautifully choreographed dance of identities and standard limits. Let us embark on this journey together.

The Indeterminate Trap

We begin, as we always do, by testing the waters. We are given the limit:

When we substitute $\theta =0$, we find that $\cos (0)=1$, so the numerator becomes $\tan (\pi )$, which is $0$. In the denominator, $\sin (0)=0$, so we get $\sin (0)$, which is also $0$.

We have arrived at the classic $\frac{0}{0}$ indeterminate form. This is not a dead end; it is an invitation to manipulate the expression into something more revealing.

The Trigonometric Bridge

Look closely at the arguments of our functions. The numerator has ${\cos }^2\theta$, and the denominator has ${\sin }^2\theta$. This mismatch is the core of the problem.

To use our standard limit theorems, we need the arguments to be consistent. We need a bridge between cosine and sine. That bridge is the fundamental identity:

By substituting this into our numerator, we transform the expression into a language of sines. Our limit now looks like this:

Distributing the $\pi$, we get $\tan (\pi -\pi {\sin }^2\theta )$. Now, we are getting somewhere!

The Allied Angle Transformation

Here is where we must be precise. We have $\tan (\pi -A)$, where $A=\pi {\sin }^2\theta$. Recall the allied angle property: $\tan (\pi -A)=-\tan A$.

This is a crucial step. Because $\pi -A$ lies in the second quadrant, the tangent function is negative. Applying this, our numerator simplifies to $-\tan (\pi {\sin }^2\theta )$.

Our limit is now:

Notice how both arguments, $\pi {\sin }^2\theta$ and $2\pi {\sin }^2\theta$, now approach $0$ as $\theta \to 0$. This is exactly what we need to invoke our standard limits.

The Symphony of Standard Limits

We know that ${\lim }_{x\to 0}\frac{\tan x}{x}=1$ and ${\lim }_{x\to 0}\frac{\sin x}{x}=1$. To force our expression into this form, we multiply and divide by the arguments:

As $\theta \to 0$, the first term approaches $1$, and the second term approaches $1$. We are left with the constant ratio:

The $\pi {\sin }^2\theta$ terms cancel out perfectly, leaving us with $-\frac{1}{2}$.

And there it is. Through careful identity substitution and the application of standard limits, we have tamed the expression.

The final answer is $-\frac{1}{2}$. Remember, in JEE Advanced, it is not just about getting the answer; it is about understanding the path. Keep practicing, keep questioning, and keep falling in love with the logic behind the math.