JEE Main 2025 — 29 January, Evening Shift. Previous Year Question.
Problem
If $\sin x+\sin^2x=1$, $x\in\left(0,\dfrac{\pi}{2}\right)$, then
$$\left(\cos^{12}x+\tan^{12}x\right)+3\left(\cos^{10}x+\tan^{10}x+\cos^{8}x+\tan^{8}x\right)+\left(\cos^{6}x+\tan^{6}x\right)$$
is equal to?
Key insight. The given condition $\sin x+\sin^2x=1$ isn’t just a starting fact to substitute once — it secretly forces $\tan x=\cos x$, which makes every “$\cos^k x + \tan^k x$” pair in the expression collapse to $2\cos^k x$. And once the whole expression is rewritten in cosines alone, its coefficients $1,3,3,1$ are a giveaway that it’s a perfect cube in disguise.
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Approach
First manipulate the given condition to discover that $\tan x=\cos x$. Use this to replace every $\tan^k x$ term in the expression with $\cos^k x$, turning the whole thing into a sum of even powers of $\cos x$ alone. Recognise the resulting coefficients as those of a binomial cube, factor accordingly, and use the original condition one more time to collapse the final bracket to $1$.
Solution
Step 1 — Derive tan x = cos x from the given condition
$$\sin x+\sin^2x=1 \implies \sin x = 1-\sin^2x = \cos^2x$$
Dividing both sides by $\cos x$:
$$\frac{\sin x}{\cos x} = \cos x \implies \tan x = \cos x$$
Step 2 — Replace every tan term with cos
Since $\tan x=\cos x$, every pair $\cos^kx+\tan^kx$ becomes $2\cos^kx$:
$$\left(\cos^{12}x+\tan^{12}x\right)+3\left(\cos^{10}x+\tan^{10}x+\cos^{8}x+\tan^{8}x\right)+\left(\cos^{6}x+\tan^{6}x\right)$$
$$= 2\cos^{12}x + 3\left(2\cos^{10}x+2\cos^{8}x\right)+2\cos^6x = 2\cos^{12}x+6\cos^{10}x+6\cos^8x+2\cos^6x$$
Step 3 — Recognise the binomial cube
Let $A=\cos^4x$ and $B=\cos^2x$. Then $A^3=\cos^{12}x$, $3A^2B=3\cos^{10}x$, $3AB^2=3\cos^8x$, $B^3=\cos^6x$ — exactly the expansion of $(A+B)^3$:
$$2\cos^{12}x+6\cos^{10}x+6\cos^8x+2\cos^6x = 2(A+B)^3 = 2\left(\cos^4x+\cos^2x\right)^3$$
Step 4 — Factor and apply the original condition
$$\cos^4x+\cos^2x = \cos^2x\left(\cos^2x+1\right)$$
But from Step 1, $\cos^2x = \sin x$, so:
$$\cos^2x(\cos^2x+1) = \sin x(\sin x + 1) = \sin x+\sin^2x = 1 \quad \text{(the given condition)}$$
Step 5 — Conclude
$$2\left(\cos^4x+\cos^2x\right)^3 = 2(1)^3 = 2$$
Answer
$$2$$
Common mistakes
- Missing the substitution $\tan x=\cos x$ entirely and trying to work with $\sin x$ and $\tan x$ separately — without this single derived fact, the expression has no clean simplification path.
- Not recognising the $1,3,3,1$ coefficient pattern. It’s tempting to try expanding or grouping the terms differently; spotting that they match a binomial cube expansion is what makes the rest of the problem collapse quickly.
Practise next
- If $\cos x+\cos^2x=1$ for $x\in\left(0,\dfrac{\pi}{2}\right)$, find $\left(\sin^{12}x+\cot^{12}x\right)+3\left(\sin^{10}x+\cot^{10}x+\sin^8x+\cot^8x\right)+\left(\sin^6x+\cot^6x\right)$, using an analogous substitution.
Show answer
$2$. From $\cos x+\cos^2x=1$ we get $\cos^2x=1-\cos x$, and also $\sin^2x=1-\cos^2x=\cos x$.
That single substitution collapses everything: $\cot^2x=\dfrac{\cos^2x}{\sin^2x}=\dfrac{1-\cos x}{\cos x}$, and since $\cos x=\tfrac{\sqrt5-1}{2}$ satisfies $\cos^2x+\cos x-1=0$, both $\sin^2x$ and $\cot^2x$ equal $\cos x$ itself.
Writing $u=\sin^2x=\cot^2x=\tfrac{\sqrt5-1}{2}$, the expression is $2\left(u^{6}+3u^{5}+3u^{4}+u^{3}\right)=2u^{3}(u+1)^{3}=2\left(u(u+1)\right)^{3}$. But $u^2+u=1$, so the bracket is $1$ and the value is $2$.
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