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Question

Find the value of x: cos(3π2+x)cos(2π+x)[cot(3π2x)+cot(2π+x)]=1

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Solution

At first we solve,
cos(3π2+x)puttingπ=180=cos(3×1802+x)=cos(270+x)=cos(36090+x)=cos(2π+(x90))=cos(x90)[cos(2π+x)=cosx]=cos((90x))=cos(90x)(cos(x)=cosx)=sinxNow,cos(2π+x)=cosx&cot(2π+x)=cotxNow,solvecot(3π2x)=cot(3×1802x)(putπ=180)=cot(270x)=cot(36090x)=cot(2π90x)=cot(2π(x+90))=cot(x+90)[cot2πx=cotx]=(tanx)[cot(90+θ)=tanθ]=tanxNowputtingvaluesinequationL.H.S=cos(3π2+x)cos(2π+x)(cot(3π2x)+cot(2π+x))=(sinx)×(cosx)×[tanx+cotx]=(sinxcosx)×[cosxsinx+sinxcosx]=(sinxcosx)×cos2x+sin2x(sinxcosx)=cos2x+sin2x=1=R.H.SHence,proved.

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