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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Trigonometric Ratios & Identities

If x+ ^2 x=1, x ∈(0, (π)/(2)) , then ( ^ 12 x+ ^ 12 x)+3 ( ^ 10 x+...

If sin⁡x+sin⁡2x=1,x∈(0,π2)\sin x+\sin ^2 x=1, x \in\left(0, \frac{\pi}{2}\right), then \left(\cos ^{12} x+\tan ^{12} x\right)+3(cos⁡10x+tan⁡10x+cos⁡8x+tan⁡8x)+(cos⁡6x+tan⁡6x)\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 :

Options

  1. A4
  2. B1
  3. C3
  4. D2

Correct answer

D. 2

Step-by-step solution

sin⁡x+sin⁡2x=1⇒sin⁡x=cos⁡2x⇒tan⁡x=cos⁡x\begin{aligned} & \sin x+\sin ^2 x=1 \\ & \Rightarrow \sin x=\cos ^2 x \Rightarrow \tan x=\cos x \end{aligned}\thereforeGivenexpressionGiven expression=2cos⁡12x+6[cos⁡10x+cos⁡8x]+2cos⁡6x=2[sin⁡6x+3sin⁡5x+3sin⁡4x+sin⁡3x]=2sin⁡3x[(sin⁡x+1)3]=2[sin⁡2x+sin⁡x]3=2\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}$
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