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

Show that sinθ+cosθsinθcosθ+sinθcosθsinθ+cosθ =2sec2θ.

Open in App
Solution

sinθ+cosθsinθcosθ+sinθcosθsinθ+cosθ

=(sinθ+cosθ)2+(sinθcosθ)2sin2θcos2θ

=sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ2sinθcosθsin2θcos2θ

=1+1sin2θcos2θ

=2(cos2θsin2θ)

=2cos2θ

=2sec2θ

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Angle and Its Measurement
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon