JEE Mathematics 2025 Question with Solution

Given: $\sin x + \sin^2 x = 1$ and $x \in \left(0, \frac{\pi}{2}\right)$.

Find: The value of

$$(\cos^2 x + \tan^2 x) + 3(\cos^4 x + \tan^4 x + \cos^4 x + \tan^4 x) + (\cos^6 x + \tan^6 x)$$

From the given relation,

$$\sin x + \sin^2 x = 1$$

Using $1 - \sin^2 x = \cos^2 x$, we get

$$\sin x = 1 - \sin^2 x = \cos^2 x$$

Hence,

$$\tan x = \frac{\sin x}{\cos x} = \frac{\cos^2 x}{\cos x} = \cos x$$

Now substitute $\tan x = \cos x$ in the expression. Then corresponding powers of $\tan x$ become the same powers of $\cos x$. The solution working concludes that the expression simplifies to

$$2[\sin^2 x + \sin^3 x + \sin^4 x]$$

Then,

$$= 2\sin^2 x (\sin x + 1)^2$$

Using the given relation again, the final value is

$$2[\sin^2 x + \sin^3 x] = 2$$

Therefore, the correct option is C.

Given: $\sin x + \sin^2 x = 1$.

Find: The required trigonometric expression.

The key observation is

$$\sin x + \sin^2 x = 1 \Rightarrow \sin x = 1 - \sin^2 x = \cos^2 x$$

So,

$$\tan x = \frac{\sin x}{\cos x} = \cos x$$

This converts the mixed expression into one involving equal cosine powers. The extracted solution states that after expansion and regrouping, the expression reduces to

$$2[\sin^2 x + \sin^3 x + \sin^4 x]$$

Now factor:

$$2[\sin^2 x + \sin^3 x + \sin^4 x] = 2\sin^2 x(1 + \sin x + \sin^2 x)$$

Using the relation $1 = \sin x + \sin^2 x$,

$$1 + \sin x + \sin^2 x = 2$$

so the expression evaluates to the value reported in the solution, namely $2$.

Therefore, the correct option is C.