Prove \cos(\frac{3\pi}{2}+x)\cos(2\pi+x)[\cot(\frac{3\pi}{2}-x)+\cot(2\pi+x)=1

Beverley Rahman

Beverley Rahman

Answered question

2022-02-25

Prove
cos(3π2+x)cos(2π+x)[cot(3π2x)+cot(2π+x)=1

Answer & Explanation

an2gi2m9gg

an2gi2m9gg

Beginner2022-02-26Added 9 answers

cos(3π2+x)=cos(2π+xπ2)=cos(xπ2)=cos(π2x)=sinx
cos(2π+x)=cosx
cot(3π2x)=cot(2π(π2+x))=cot(π2+x)
=(tanx)=tanx
cot(2π+x)=cotx
LHS=(sinxcosx)(tanx+cotx)
=(sinxcosx)(sinxcosx+cosxsinx)
=sin2x+cos2x=1
litoshypinaylh4

litoshypinaylh4

Beginner2022-02-27Added 6 answers

cos(3π2+x)=cos(π+x+π2)=cos(π2+x)=+sinx
cos(2π+x)=cos(x)
cot((3π2)x)=cot(π(xπ2))
=cot(xπ2)
=cot(π2x)=tan(x)
cot(2π+x)=cot(x)
So:
[sin(x)×cos(x)]×[tan(x)+cot(x)]=sin2(x)+cos2(x)=1

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