Proving \(\displaystyle{\frac{{{\cos{{2}}}{x}}}{{{1}+{\sin{{2}}}{x}}}}={\sec{{2}}}{x}-{\tan{{2}}}{x}\) I tried \(\displaystyle{\frac{{{1}-{{\sin}^{{2}}{x}}}}{{{1}+{2}{\sin{{x}}}{\cos{{x}}}}}}-{\frac{{{1}-{{\cos}^{{2}}{x}}}}{{{1}+{2}{\sin{{x}}}{\cos{{x}}}}}}\) but I couldn't simplify.

Pizzadililehz

Pizzadililehz

Answered question

2022-03-28

Proving cos2x1+sin2x=sec2xtan2x
I tried
1sin2x1+2sinxcosx1cos2x1+2sinxcosx
but I couldn't simplify. I start with LHS.

Answer & Explanation

umgebautv6v2

umgebautv6v2

Beginner2022-03-29Added 10 answers

Use that
1sin2(2x)=cos2(2x)
Abdullah Avery

Abdullah Avery

Beginner2022-03-30Added 19 answers

(1+sin2xcos2x)1=(sec2x+tan2x)1=
=sec2xtan2xsec22xtan22x=
=sec2xtan2x

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