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If tan^2B=sin^2A-cos^2A then prove that tan^2A=sin^2B-cos^B
Dear Studentyou did mistake in typing question ,it should be like thiscos²A - sin²A = tan²B , then prove , cos²B - sin²B = tan²AGiven, 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 = tan²AHence provedThanks
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