How to solve \(\displaystyle{\frac{{{2}{\tan{{x}}}}}{{{1}-{\left({\tan{{x}}}\right)}^{{2}}}}}={\left({\sin{{2}}}{x}\right)}^{{2}}\)

Nathen Peterson

Nathen Peterson

Answered question

2022-03-31

How to solve
2tanx1(tanx)2=(sin2x)2

Answer & Explanation

ron4d3ozip7

ron4d3ozip7

Beginner2022-04-01Added 9 answers

Using Double Angle formula for tangent,
tan2x=sin22x
sin2x(1sin2xcos2x)=0
Now sin2xcos2x=sin4x212
Now sin2x=02x=nπ where n is any integer
jmroberts70pbo2

jmroberts70pbo2

Beginner2022-04-02Added 10 answers

2tanx1tan2x=(sin2x)2
tan2x=sin22x
sin2xcos2x=sin22x
sin2xcos2xsin22x=0
sin2xcos2x(1sin2xcos2x)=0
either sin2x=0 or 1sin2xcos2x=0
sin2x=0 gives x=nπ2
1sin2xcos2x=0sin4x=2 which is not possible as sin cannot be greater than 1.

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