Sin^2A-cos^2A.cos2B=sin^2B-cos^2B.cos2A prove it
Answers
Answered by
20
sin^2A×cos2B - sin^2B×cos2A
= sin^2 A ( cos^2 B - sin^2 B) - sin^2 B ( cos^2 A - sin^2 A)
= sin^2 A cos^2 B - sin^2 A sin^2 B - sin^2 B cos^2 A + sin^2 B sin^2 A
= sin^2 A cos^2 B - sin^2 B cos^2 A
= (1-cos^2 A) cos^2 B - (1-cos^2 B) cos^2 A
= cos^2 B - cos^2 A cos^2 B - cos^2 A + cos^2B cos^2 A
= cos^2 B - cos^2 A
= sin^2 A ( cos^2 B - sin^2 B) - sin^2 B ( cos^2 A - sin^2 A)
= sin^2 A cos^2 B - sin^2 A sin^2 B - sin^2 B cos^2 A + sin^2 B sin^2 A
= sin^2 A cos^2 B - sin^2 B cos^2 A
= (1-cos^2 A) cos^2 B - (1-cos^2 B) cos^2 A
= cos^2 B - cos^2 A cos^2 B - cos^2 A + cos^2B cos^2 A
= cos^2 B - cos^2 A
Answered by
22
Answer:
Step-by-step explanation:
Attachments:
Similar questions