Question

sin4 theta+sin2 theta/1+cos2 theta +cos4 theta =tan 2 theta

Answer 1

Check the attachment

Answer 2

(Sin4θ  + Sin2θ)/(1  + Cos2θ  +Cos4θ)  = Tan2θ

Step-by-step explanation:

(Sin4θ  + Sin2θ)/(1  + Cos2θ  +Cos4θ)  = Tan2θ

LHS = (Sin4θ  + Sin2θ)/(1  + Cos2θ  +Cos4θ)

Using  Sin2x = 2SinxCosx  => Sin4θ = 2Sin2θCos2θ

& Cos2x = 2Cos²x - 1   => Cos4θ = 2Cos²2θ  - 1

= (2Sin2θCos2θ + Sin2θ)/(1 + Cos2θ  + 2Cos²2θ  - 1)

= Sin2θ(2Cos2θ + 1) /(Cos2θ  + 2Cos²2θ)

= Sin2θ(2Cos2θ + 1) /Cos2θ(1   + 2Cos2θ)

= Sin2θ(1 + 2Cos2θ) /Cos2θ(1   + 2Cos2θ)

Cancelling 1   + 2Cos2θ

= Sin2θ /Cos2θ

= Tan2θ

= RHS

QED

Proved

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