Question

PROVE THAT THE FOLLOWING QUESTION.

Answer 1

cot(90-θ) sin(90-θ)/sin θ + cot 40/tan50 -( cos^2(20) + cos^2(70))=1

Proof

We know that,

cot(9-θ)=tan θ

sin(90-θ)=cos θ

tan 50=cot(90-50)

cos^2(70)=sin^2(90-70)

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tanθ*cosθ/sin θ+cot40/cot(90-50)-( cos^2(20) + sin^2(90-70))------------>a

tan θ=sinθ/cos θ (put this in eq a)

sin θ/cos θ*cosθ/sinθ+cot40/cot40-( cos^2(20) + sin^2(20))----------------->b

sin θ/cos θ*cosθ/sinθ=1

cot40/cot40=1

cos^2(θ) + sin^2(θ)=1,

so ,

( cos^2(20) + sin^2(20))=1

1+1-1=1

cot(90-θ) sin(90-θ)/sin θ + cot 40/tan50 -( cos^2(20) + cos^2(70))=1