wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Prove that sin2θcos2θ+cos2θsin2θ=sec2θcosec2θ2

Open in App
Solution

LHS=sin2θcos2θ+cos2θsin2θ

=sin4θ+cos4θsin2θ.cos2θ

We know that,
a2+b2+2ab=(a+b)2 (a+b)22ab=a2+b2

=(sin2θ+cos2θ)22sin2θ.cos2θsin2θ.cos2θ

=12sin2θ.cos2θsin2θ.cos2θ

=1sin2θ.cos2θ2

=sec2θ(cosec2θ)2 (1+cot2θ=cosec2θ)

=sec2θ(1+cot2θ)2 (cosθ=1secθ)

=sec2θ+sec2θ.cos2θsin2θ2

=sec2θ+cosec2θ2

=RHS

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Relation between Trigonometric Ratios
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon