JEE Main 2019TrigonometryTrigonometric Ratio and IdentitesEasyQuestionIf 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)(\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)(cos12x+tan12x)+3(cos10x+tan10x+cos8x+tan8x)+(cos6x+tan6x) is equal to:OptionsA3B4C2D1Check AnswerHide SolutionSolutionsinx+sin2x=1⇒sinx=cos2x⇒tanx=cosx∴ Given expression =2cos12x+6[cos10x+cos8x]+2cos6x=2[sin6x+3sin5x+3sin4x+sin3x]=2sin3x[(sinx+1)3]=2[sin2x+sinx]3=2\begin{aligned} &\begin{aligned} & \sin x+\sin ^2 x=1 \\ & \Rightarrow \sin x=\cos ^2 x \Rightarrow \tan x=\cos x \end{aligned}\\ &\therefore \text { Given expression }\\ &\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} \end{aligned}sinx+sin2x=1⇒sinx=cos2x⇒tanx=cosx∴ Given expression =2cos12x+6[cos10x+cos8x]+2cos6x=2[sin6x+3sin5x+3sin4x+sin3x]=2sin3x[(sinx+1)3]=2[sin2x+sinx]3=2