Animated Solution for Mathematics - Trigonometry: If $sinx+si{n}^{2}x=1,x\in (0,\frac{\pi }{2})$, then $(co{s}^{12}x+ta{n}^{12}x)+3(co{s}^{10}x+ta{n}^{10}x+co{s}^{8}x+ta{n}^{8}x)+(co{s}^{6}x+ta{n}^{6}x)$ is equal to:

Imagine you are sitting in the examination hall, the clock is ticking, and you are faced with this trigonometric behemoth:

Your first instinct might be panic. It looks like a chaotic mess of high-power trigonometric functions.

But here is the secret of the JEE Advanced: whenever you see a problem that looks like a nightmare, it is almost certainly hiding a beautiful, elegant symmetry. Let us peel back the layers together.

Every great mystery has a key. Our key is the very first line: $\sin x+{\sin }^{2}x=1$.

This is not just an equation; it is a transformation waiting to happen. If we rearrange this, we get $\sin x=1-{\sin }^{2}x$.

Now, pause. What is $1-{\sin }^{2}x$? It is the fundamental identity ${\cos }^{2}x$.

So, in one swift move, we have established that $\sin x={\cos }^{2}x$. This is our bridge. It tells us that wherever we see a ${\cos }^{2}x$, we can replace it with $\sin x$.

Now, what about those tangent terms? They look intimidating, but they are just $\sin x$ in disguise.

By definition, ${\tan }^{2}x=\frac{{\sin }^{2}x}{{\cos }^{2}x}$. Since we just proved that ${\cos }^{2}x=\sin x$, we can substitute that into the denominator.

The expression becomes:

One $\sin x$ cancels out, and we are left with the stunningly simple result: ${\tan }^{2}x=\sin x$. Now we have both ${\cos }^{2}x$ and ${\tan }^{2}x$ equal to $\sin x$.

Look back at that massive expression. Do you see the coefficients? $1,3,3,1$.

These are not random numbers. They are the binomial coefficients from Pascal's Triangle, the signature of the expansion $(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$.

If we group the cosine terms and the tangent terms, we get:

This is a perfect cube! For the cosine group, if we set $a={\cos }^{4}x$ and $b={\cos }^{2}x$, the expression becomes $({\cos }^{4}x+{\cos }^{2}x{)}^3$. The same logic applies to the tangent group, yielding $({\tan }^{4}x+{\tan }^{2}x{)}^3$.

Now, let us bring it all home. We know ${\cos }^{2}x=\sin x$, which implies ${\cos }^{4}x={\sin }^{2}x$.

Substituting these into our cosine group, we get $({\sin }^{2}x+\sin x{)}^3$. But wait—the problem told us at the very beginning that $\sin x+{\sin }^{2}x=1$.

So, the cosine group is just $1^3$, which is $1$. The exact same logic applies to the tangent group:

Adding them together, $1+1=2$. The monster has been tamed.

The final answer is 2. It was never a complex polynomial; it was just a beautiful, hidden identity waiting for you to find it.