Question

pls solve this question

Answer 1

Observe that

sin4θ+cos4θ=1−2sin2θcos2θ⟺sin4θ+cos4θ+2sin2θcos2θ=1sin4⁡θ+cos4⁡θ=1−2sin2⁡θcos2⁡θ⟺sin4⁡θ+cos4⁡θ+2sin2⁡θcos2⁡θ=1

which is always true since

sin4θ+cos4θ+2sin2θcos2θ=(sin2θ+cos2θ)2=12=1sin4⁡θ+cos4⁡θ+2sin2⁡θcos2⁡θ=(sin2⁡θ+cos2⁡θ)2=12=1

so the identity is proved.

I hope it will help you dear...

Answer 2

sin4θ+cos4θ=1−2sin2θcos2θ⟺sin4θ+cos4θ+2sin2θcos2θ=1sin4⁡θ+cos4⁡θ=1−2sin2⁡θcos2⁡θ⟺sin4⁡θ+cos4⁡θ+2sin2⁡θcos2⁡θ=1

which is always true since

sin4θ+cos4θ+2sin2θcos2θ=(sin2θ+cos2θ)2=12=1sin4⁡θ+cos4⁡θ+2sin2⁡θcos2⁡θ=(sin2⁡θ+cos2⁡θ)2=12=1