If 10 sin⁴ θ + 15 cos⁴ θ = 6, then the value of (27 cosec⁶ θ + 8 sec⁶ θ) / (16 sec⁸ θ) is:

Q: Wait, why do we convert everything to $\sin^{2} \theta$? Why not just solve for $\theta$ directly?

Great question! In JEE problems, we rarely need the actual angle $\theta$. Solving for $\theta$ would involve inverse trigonometric functions, which is messy and unnecessary. By treating $\sin^{2} \theta$ as a single variable, say $x$, we transform a complex trigonometric equation into a simple algebraic quadratic equation. Always look for the 'hidden variable'!

Q: Is it always a perfect square in these types of problems?

Not always, but in competitive exams like JEE, the coefficients are often carefully chosen to lead to a perfect square or a factorable quadratic. If you don't see a perfect square, don't panic—just use the standard quadratic formula $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.

Q: Why did the powers of 6 and 8 in the target expression simplify so nicely?

That is the beauty of the problem design! The coefficients $27$ and $8$ are actually $3^{3}$ and $2^{3}$. When you substitute $\sin^{2} \theta = \frac{3}{5}$ and $\cos^{2} \theta = \frac{2}{5}$, the denominators of the fractions cancel out perfectly with these coefficients. It's a reward for your algebraic persistence.

Q: Can I use $\sin^{2} \theta + \cos^{2} \theta = 1$ to simplify the target expression before substituting?

Absolutely! In fact, that is a more advanced approach. If you can manipulate the expression to look like a function of $\tan \theta$ or $\cot \theta$, it might save you time. However, direct substitution is often safer and less prone to algebraic errors under exam pressure.

Q: What if I get a negative value for $\sin^{2} \theta$?

That would be a red flag! Since $\sin^{2} \theta$ represents the square of a real number, it must be non-negative. If you get a negative result, re-check your expansion of $(1 - \sin^{2} \theta)^{2}$. It is the most common place to make a sign error.

Question

If $10 \sin^{4} 	heta + 15 \cos^{4} 	heta = 6$, then the value of $\frac{27 \operatorname{cosec}^{6} 	heta + 8 \sec^{6} 	heta}{16 \sec^{8} 	heta}$ is:

Options:

(A) $\frac{2}{5}$

(B) $\frac{3}{4}$

(C) $\frac{3}{5}$

(D) $\frac{1}{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE excellence. Today, we aren't just solving a trigonometric equation; we are peeling back the layers of a mathematical onion to reveal the elegant core hidden within.

When you first look at the equation $10 \sin^{4} 	heta + 15 \cos^{4} 	heta = 6$, it might seem like a daunting landscape of powers and ratios. In the world of JEE, complexity is often just a mask for symmetry.

### The Algebraic Transformation

Our first mission is to simplify the battlefield. We have two different trigonometric functions, $\sin 	heta$ and $\cos 	heta$, so let us unify them using the fundamental identity $\cos^{2} 	heta = 1 - \sin^{2} 	heta$.

By substituting this into our equation, we shift the entire problem into the realm of $\sin^{2} 	heta$. Let us define a new variable $x = \sin^{2} 	heta$. Our equation transforms into:

$$10x^{2} + 15(1 - x)^{2} = 6$$

Expanding the term $(1 - x)^{2}$ gives us $1 + x^{2} - 2x$. Distributing the $15$ and combining like terms, we arrive at the quadratic equation:

$$25x^{2} - 30x + 9 = 0$$

### The Beauty of the Perfect Square

Look closely at $25x^{2} - 30x + 9 = 0$. It is the classic expansion of $(5x - 3)^{2} = 0$. This is the hallmark of a well-crafted problem, telling us that $5x = 3$, or simply:

$$\sin^{2} 	heta = \frac{3}{5}$$

With this golden key in hand, finding $\cos^{2} 	heta$ becomes trivial:

$$\cos^{2} 	heta = 1 - \frac{3}{5} = \frac{2}{5}$$

We have successfully decoded the relationship between these two functions. We are no longer guessing; we are calculating with precision.

### The Final Ascent

Now, we turn our attention to the target expression:

$$\frac{27 \operatorname{cosec}^{6} 	heta + 8 \sec^{6} 	heta}{16 \sec^{8} 	heta}$$

Since $\operatorname{cosec}^{2} 	heta = \frac{1}{\sin^{2} 	heta} = \frac{5}{3}$ and $\sec^{2} 	heta = \frac{1}{\cos^{2} 	heta} = \frac{5}{2}$, we can rewrite the expression in terms of these squares:

$$\frac{27(\operatorname{cosec}^{2} 	heta)^{3} + 8(\sec^{2} 	heta)^{3}}{16(\sec^{2} 	heta)^{4}}$$

Substitute our values: the numerator becomes $27(\frac{5}{3})^{3} + 8(\frac{5}{2})^{3}$. Notice how the $27$ in the numerator cancels the $3^{3}$ in the denominator, and the $8$ cancels the $2^{3}$. We are left with $125 + 125 = 250$.

Finally, the denominator is calculated as:

$$16\left(\frac{5}{2}ight)^{4} = 16 	imes \frac{625}{16} = 625$$

Our final result is $\frac{250}{625}$. Dividing both by $125$, we arrive at the elegant answer:

$$\mathbf{\frac{2}{5}}$$

### Reflection

We started with a complex trigonometric expression and, through logical substitution and algebraic factorization, reduced it to a simple fraction. This is the essence of JEE mathematics.

It is not about memorizing formulas; it is about recognizing patterns and having the patience to let the algebra reveal the truth. You have the tools, you have the logic, and now, you have the victory.

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

Select Answer:

Visualized Solution

Analyze the Given Equation

Substitute $cos^{2} θ$ using Identity

Expand the Square Term

Distribute and Simplify

Form the Quadratic Equation

Factorize as a Perfect Square

Solve for $sin^{2} θ$ and $cos^{2} θ$

Find Reciprocal Ratios

Substitute into the Target Expression

Simplify the Numerator

Simplify the Denominator

Final Calculation

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Algebraic Transformation

The Beauty of the Perfect Square

The Final Ascent

Reflection

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 10sin4θ+15cos4θ=6, then the value of 16sec8θ27cosec6θ+8sec6θ​ is: