$\begin{aligned}
& \sin x+\sin ^2 x=1 \
& \Rightarrow \sin x=\cos ^2 x \Rightarrow \tan x=\cos x
\end{aligned}\thereforeGivenexpression\begin{aligned}
& =2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^8 x\right]+2 \cos ^6 x \
& =2\left[\sin ^6 x+3 \sin ^5 x+3 \sin ^4 x+\sin ^3 x\right] \
& =2 \sin ^3 x\left[(\sin x+1)^3\right] \
& =2\left[\sin ^2 x+\sin x\right]^3 \
& =2
\end{aligned}$
If sin x+sin ^2 x=1, x ∈ (0, π 2 ), then (cos ^12 x+tan ^12 x )+3 (cos ^10 x+tan ^10 x+cos ^8 x+tan ^8 x )+ (cos ^6 x+tan ^6 x ) is equal to :
JEE Main 2025 — Mathematics Trigonometry
If sinx+sin2x=1,x∈(0,2π), then (cos12x+tan12x)+3(cos10x+tan10x+cos8x+tan8x)+(cos6x+tan6x) is equal to :
Held on 29 Jan 2025 · Verified 6 Jul 2026.
Options
- A
4
- B
1
- C
3
- D
2
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