If cos^2A- sin^2A=tan^2B ,prove that tan^2A =cos^2B-sin^2B

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cos²A - sin²A = tan²B , then prove , cos²B - sin²B = tan²A

Given, cos²A - sin²A = tan²B

⇒cos²A - ( 1 - cos²A ) = tan²B [∵ sin²Ф = 1 - cos²Ф ]

⇒ cos²A - 1 + cos²A = tan²B

⇒ 2cos²A - 1 = tan²B

⇒ 2cos²A = 1 + tan²B

⇒ 2cos²A = sec²B [ ∵ sec²Ф = 1 + tan²Ф ]

⇒2cos²A.cos²B =1 [ ∵ secФ = 1/cosФ ]

⇒2cos²B = sec²A

= cos²B + cos²B = 1 + tan²A

⇒ cos²B + cos²B - 1 = tan²A

⇒cos²B - (1 - cos²B) = tan²A

⇒ cos²B - sin²B = tan²A

Hence proved//

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