Animated Solution for Mathematics - Trigonometry: Let $S=\left\{\theta \in [-\pi ,\pi ]-\left\{\pm \frac{\pi }{2}\right\}:\sin \theta \tan \theta +\tan \theta =\sin 2\theta \right\}$. If $T={\sum }_{\theta \in S}\cos 2\theta$, then $T+n(S)$ is equal to

Welcome, future engineer! Today, we are not just solving an equation; we are embarking on a journey through the landscape of trigonometry. Imagine standing at the origin of the unit circle, looking out at the vast expanse of angles from $-\pi$ to $\pi$.

Our equation, $\sin \theta \tan \theta +\tan \theta =\sin 2\theta$, is a complex puzzle. The first thing that should catch your eye is the domain constraint: $\theta \in [-\pi ,\pi ]-\{\pm \frac{\pi }{2}\}$.

These points are excluded because at $\pm \frac{\pi }{2}$, the tangent function explodes to infinity. It is a 'forbidden zone' where the math breaks down; we must respect this boundary to avoid chasing ghosts.

Now, look at the equation itself. We see $\tan \theta$ on the left. A common mistake is to divide by it immediately, but remember: if you divide by a variable, you might lose the very solutions that make that variable zero.

Instead, we factor the expression:

This is our 'Algebraic Fork' in the road. Case 1 is $\tan \theta =0$. This occurs when $\sin \theta =0$, which yields the solutions $\theta =-\pi ,0,\pi$. These are our first three soldiers in the set $S$.

For Case 2, we assume $\tan \theta eq 0$ and safely divide. We use the identity $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$. Substituting this, we get:

Since $1+{\tan }^2\theta ={\sec }^2\theta$, the right side simplifies to $2{\cos }^2\theta$. We are now in the realm of a quadratic equation:

Rearranging this, we arrive at:

Factoring this quadratic gives $(2\sin \theta -1)(\sin \theta +1)=0$. We have two potential roots: $\sin \theta =1/2$ and $\sin \theta =-1$.

Check the domain again. $\sin \theta =-1$ implies $\theta =-\pi /2$, which is in our forbidden zone, so we must reject it. The only valid roots from this case are $\sin \theta =1/2$, which gives $\theta =\pi /6$ and $5\pi /6$.

We gather our set $S=\{-\pi ,0,\pi ,\pi /6,5\pi /6\}$. The number of elements $n(S)$ is 5.

To find $T$, we sum $\cos 2\theta$ for each element:

Adding $T+n(S)$ gives us $4+5=9$. You have conquered the problem! Keep this mindset of checking constraints and factoring carefully, and no JEE problem will ever stand in your way. The final answer is 9.