Prove that: cos²A +sin²A.cos2B = cos²B +sin²B.cos2A

Prove that: cos²A +sin²A.cos2B = cos²B +sin²B.cos2A

This is a class 10 Question From Trigonometric Identities chapter of Unit Trigonometry. All the steps for the solutions are mentioned as hint. If that's hard for you to navigate, you can always visit the facebook link given at the end of every posts. 

Solution:

Taking LHS;

= cos²A +sin²A.cos2B

[ cos2B = cos²B -sin²B ]

= cos²A +sin²A.(cos²B -sin²B)

= cos²A +cos²B.sin²A -sin²A.sin²B

[ sin²A = 1- cos²A ]

= cos²A +cos²B.(1-cos²A) -sin²A.sin²B

= cos²A +cos²B - cos²A.cos²B - sin²A.sin²B

= cos²B +cos²A - cos²A.cos²B - sin²A.sin²B

= cos²B +cos²A (1 - cos²B) -sin²A.sin²B

[ 1- cos²B = sin²B ]

= cos²B +cos²A.sin²B -sin²A.sin²B

= cos²B +sin²B.(cos²A -sin²A)

[ cos²A - sin²A = cos 2A ]

= cos²B +sin²B.cos2A

= RHS


Here is the Facebook link to the solution of this question in image. 

Related Notes:

Link: Introduction To Trigonometry
Link: Values of Trigonometric Ratios
Link: Compound Angles

See all the solutions of Trigonometric Identities in this page. 

Question: Prove that: cos²A +sin²A.cos2B = cos²B +sin²B.cos2A | SciPiPupil


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