If sin x+sin 2 x=1, x in (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

Mathematics · Trigonometry Ratios and Identities

JEE Main 2025 — 29 January, Evening Shift — Question 43

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

Option A:
4
Option B:
3
Option C:
2
Correct
Option D:
1

Answer: C

Step-by-step solution

$\quad \sin x+\sin ^{2} x=1$

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

$\therefore$ Given expression $=2 \cos ^{12} x+6\left[\cos ^{10} x+\cos ^{8} x\right]+2 \cos ^{6} x$

$=2\left[\sin ^{6} x+3 \sin ^{5} x+3 \sin ^{4} x+\sin ^{3} x\right]$

$=2 \sin ^{3} x\left[(\sin x+1)^{3}\right]$

$=2\left[\sin ^{2} x+\sin x\right]^{3}$

$=2$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Mathematics
Chapter: Trigonometry Ratios and Identities
Topic: Conditional identities in Trigonometric Ratios