Asked by Kate
                Prove each identity:
a) 1-cos^2x=tan^2xcos^2x
b) cos^2x + 2sin^2x-1 = sin^2x
I also tried a question on my own:
tan^2x = (1 – cos^2x)/cos^2x
R.S.= sin^2x/cos^2x
I know that the Pythagorean for that is sin^2x + cos^2x
That's all I could do.
            
        a) 1-cos^2x=tan^2xcos^2x
b) cos^2x + 2sin^2x-1 = sin^2x
I also tried a question on my own:
tan^2x = (1 – cos^2x)/cos^2x
R.S.= sin^2x/cos^2x
I know that the Pythagorean for that is sin^2x + cos^2x
That's all I could do.
Answers
                    Answered by
            MathMate
            
    a)
Given tan(θ)=sin(θ)/cos(θ)
substitute tan(θ) by sin(θ)/cos(θ) on the RHS and see what you get.
b) hint:
cos(2θ)=cos²(θ)-sin²(θ)
    
Given tan(θ)=sin(θ)/cos(θ)
substitute tan(θ) by sin(θ)/cos(θ) on the RHS and see what you get.
b) hint:
cos(2θ)=cos²(θ)-sin²(θ)
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