Animated Solution for Mathematics - Trigonometry: If $\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5}$, then

Welcome, future engineer. Today, we are going to dissect a problem that, at first glance, might seem like a chaotic mess of powers and fractions. You see $\frac{{\sin }^{4}x}{2}+\frac{{\cos }^{4}x}{3}=\frac{1}{5}$ and your instinct might be to panic.

But in the world of JEE Advanced, we don't panic; we analyze. We look for the hidden structure beneath the surface. This problem is a masterclass in how trigonometry and algebra dance together.

The Strategy of Reduction

The first thing we must do is simplify our landscape. We have two variables—$\sin x$ and $\cos x$—and they are raised to the fourth power. This is the 'trap' of the problem.

If you try to solve this using double-angle formulas or complex identities, you will find yourself in a labyrinth of terms. Instead, we use the most powerful tool in our trigonometric toolkit: the fundamental identity ${\sin }^{2}x+{\cos }^{2}x=1$.

By expressing everything in terms of ${\sin }^{2}x$, we reduce the problem from a trigonometric nightmare to a simple algebraic quadratic. We rewrite ${\cos }^{4}x$ as $({\cos }^{2}x{)}^2$, which is $(1-{\sin }^{2}x{)}^2$. Now, our equation looks like this:

The Algebraic Transformation

Now, let us expand that numerator. Remember the identity $(a-b{)}^2=a^2-2ab+b^2$. Applying this to $(1-{\sin }^{2}x{)}^2$, we get $1-2{\sin }^{2}x+{\sin }^{4}x$.

Our equation now reads:

Fractions are the enemies of clarity. To clear them, we look for the Least Common Multiple of the denominators 2, 3, and 5, which is 30. Multiplying the entire equation by 30, we get:

Distributing the 10 and combining like terms, we arrive at a beautiful, clean quadratic equation:

The "Aha!" Moment

Look at that equation again. $25{\sin }^{4}x-20{\sin }^{2}x+4=0$. Does it look familiar? It is a perfect square!

If we let $u={\sin }^{2}x$, we have $25u^2-20u+4=0$. This is $(5u-2{)}^2=0$.

This is the moment where the problem collapses into simplicity. Because the square of an expression is zero, the expression itself must be zero. Thus, $5{\sin }^{2}x-2=0$, which gives us the elegant result:

From here, finding ${\cos }^{2}x$ is trivial. Since ${\cos }^{2}x=1-{\sin }^{2}x$, we have ${\cos }^{2}x=1-\frac{2}{5}=\frac{3}{5}$. We have unlocked the core values of the problem.

The Final Verification

The question asks for ${\tan }^{2}x$ and the value of $\frac{{\sin }^{8}x}{8}+\frac{{\cos }^{8}x}{27}$.

For ${\tan }^{2}x$, we simply take the ratio:

This confirms Option A. Now, for the second part, we use our values of ${\sin }^{2}x$ and ${\cos }^{2}x$ raised to the fourth power:

Calculating these powers, we get $\frac{16/625}{8}+\frac{81/625}{27}$, which simplifies to $\frac{2}{625}+\frac{3}{625}=\frac{5}{625}=\frac{1}{125}$. This confirms Option B.

We started with a daunting equation and ended with a clear, logical path. The lesson here is simple: never let the complexity of an expression intimidate you.

Look for the underlying algebraic structure, reduce your variables, and trust your identities. You have the tools; you just need to apply them with confidence. Keep practicing, and soon, these patterns will become second nature to you.