JEE Mathematics 2025 Question with Solution

Given: $10 \sin^4 \theta + 15 \cos^4 \theta = 6$

Find: $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$

Let $u = \sin^2 \theta$. Then $\cos^2 \theta = 1-u$, so

$$10u^2 + 15(1-u)^2 = 6$$

Expanding,

$$10u^2 + 15(1-2u+u^2) = 6$$ $$10u^2 + 15 - 30u + 15u^2 = 6$$ $$25u^2 - 30u + 9 = 0$$

Now solve the quadratic:

$$u = \frac{30 \pm \sqrt{900-900}}{50} = \frac{30}{50} = \frac{3}{5}$$

Hence,

$$\sin^2 \theta = \frac{3}{5}, \qquad \cos^2 \theta = \frac{2}{5}$$

Therefore,

$$\csc^2 \theta = \frac{5}{3}, \qquad \sec^2 \theta = \frac{5}{2}$$

So,

$$\csc^6 \theta = \left(\frac{5}{3}\right)^3 = \frac{125}{27}, \qquad \sec^6 \theta = \left(\frac{5}{2}\right)^3 = \frac{125}{8}, \qquad \sec^8 \theta = \left(\frac{5}{2}\right)^4 = \frac{625}{16}$$

Substitute these values into the expression:

$$\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta} = \frac{27 \cdot \frac{125}{27} + 8 \cdot \frac{125}{8}}{16 \cdot \frac{625}{16}}$$ $$= \frac{125 + 125}{625} = \frac{250}{625} = \frac{2}{5}$$

Therefore, the correct option is A, that is $\frac{2}{5}$.

Given: $10 \sin^4 \theta + 15 \cos^4 \theta = 6$

Find: $\frac{27 \csc^6 \theta + 8 \sec^6 \theta}{16 \sec^8 \theta}$

Observe that the given relation involves only $\sin^4 \theta$ and $\cos^4 \theta$, so setting $x = \sin^2 \theta$ converts it into a quadratic in one variable.

$$10x^2 + 15(1-x)^2 = 6$$ $$25x^2 - 30x + 9 = 0$$

Since the discriminant is zero, the value of $x$ is unique:

$$x = \frac{3}{5}$$

Thus,

$$\sin^2 \theta = \frac{3}{5}, \qquad \cos^2 \theta = \frac{2}{5}$$

Immediately,

$$\csc^6 \theta = \left(\frac{5}{3}\right)^3, \qquad \sec^6 \theta = \left(\frac{5}{2}\right)^3, \qquad \sec^8 \theta = \left(\frac{5}{2}\right)^4$$

Now substitute:

$$\frac{27 \left(\frac{5}{3}\right)^3 + 8 \left(\frac{5}{2}\right)^3}{16 \left(\frac{5}{2}\right)^4} = \frac{125 + 125}{625} = \frac{2}{5}$$

Therefore, the correct option is A.