Animated Solution for Mathematics - Trigonometry: If $10{\sin }^4\theta +15{\cos }^4\theta =6$, then the value of $\frac{27{\mathrm{cosec}}^6\theta +8{\sec }^6\theta }{16{\sec }^8\theta }$ is:

A Journey into Trigonometric Elegance

Welcome, future engineer. When you first look at the equation $10{\sin }^4\theta +15{\cos }^4\theta =6$, it is natural to feel a surge of intimidation. You see powers of four, a mix of sine and cosine, and a target expression that looks like a nightmare of secants and cosecants.

But here is the secret of the JEE Advanced: complexity is often just a mask for a simple, elegant truth waiting to be revealed. Let us peel back that mask together.

The Algebraic Transformation

The first step in any great problem is to simplify the landscape. We have an equation with both sine and cosine. We know the fundamental identity ${\sin }^2\theta +{\cos }^2\theta =1$, which implies ${\cos }^2\theta =1-{\sin }^2\theta$.

If we square this, we get ${\cos }^4\theta =(1-{\sin }^2\theta {)}^2$. By substituting this into our original equation, we shift the problem from the realm of trigonometry into the realm of algebra.

Let us define a variable $t={\sin }^2\theta$. Suddenly, the equation $10{\sin }^4\theta +15{\cos }^4\theta =6$ transforms into:

This is much friendlier, isn't it?

The Quadratic Reveal

Now, we expand the term $(1-t{)}^2$ using the identity $(a-b{)}^2=a^2-2ab+b^2$. This gives us $10t^2+15(1+t^2-2t)=6$.

Distributing the $15$ leads us to $10t^2+15+15t^2-30t=6$. Combining the like terms, we get $25t^2-30t+15=6$.

Subtracting $6$ from both sides, we arrive at the beautiful quadratic equation:

Look closely at this equation. It is a perfect square! $25t^2$ is $(5t{)}^2$, $9$ is $3^2$, and $-30t$ is $-2(5t)(3)$.

Thus, the equation collapses into $(5t-3{)}^2=0$. This tells us that $5t=3$, or $t=\frac{3}{5}$.

Since $t={\sin }^2\theta$, we have found our core value: ${\sin }^2\theta =\frac{3}{5}$. Consequently, ${\cos }^2\theta =1-\frac{3}{5}=\frac{2}{5}$.

The Elegant Collapse

Now, we turn our attention to the target expression:

We know that ${\mathrm{cosec}}^2\theta =\frac{1}{{\sin }^2\theta }=\frac{5}{3}$ and ${\sec }^2\theta =\frac{1}{{\cos }^2\theta }=\frac{5}{2}$. We can rewrite the powers of $6$ and $8$ in terms of these squares.

The numerator becomes $27({\mathrm{cosec}}^2\theta {)}^3+8({\sec }^2\theta {)}^3$, and the denominator becomes $16({\sec }^2\theta {)}^4$. Substituting our values, the numerator is:

Notice the magic? The $27$ and $8$ cancel out perfectly, leaving us with $125+125=250$. The denominator is:

Our final fraction is $\frac{250}{625}$. Dividing both by $125$, we get the final result:

This is the beauty of the JEE Advanced—a problem that starts with a terrifying expression and ends with a simple, clean fraction. Keep this mindset: when you see a complex expression, look for the substitution that makes it simple. You have the tools; now go forth and solve.