JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumTrigonometry118× since 2002If sinx+sin2x=1\sin x + \sin^2 x = 1sinx+sin2x=1, x∈(0,π2)x \in \left(0, \frac{\pi}{2}\right)x∈(0,2π), then (cos12x+tan12x)+3(cos10x+tan10x+cos8x+tan8x)+(cos6x+tan6x)\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)(cos12x+tan12x)+3(cos10x+tan10x+cos8x+tan8x)+(cos6x+tan6x) is equal to:AOption A. 4BOption B. 1COption C. 3DOption D. 2CheckSkip · show solution
