InterviewSolution
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InterviewSolution
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Prove the following identities:`tan^2A-tan^2B=(cos^2B-cos^2A)/(cos^2Bcos^2A)=(sin^2A-sin^2B)/(cos^2Acos^2B)``(sinA-sinB)/(cosA+cosB)+(cosA-cosB)/(sinA+sinB)=0` |
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Answer» We have ` (tan^(2)A - tan^(2)B) = (sin^(2)A)/(cos^(2)A) - (sin^(2)B)/(cos^(2)B) ` ` = (sin^(2)A cos^(2)B - cos^(2)A sin^(2)B)/(cos^(2)A cos^(2)B) ` ` = (sin^(2)A(1- sin^(2)B)-(1- sin^(2)A)sin^(2)B )/(cos^(2)A cos^(2)B) ` ` = ((sin^(2)A - sin^(2)B))/(cos^(2)A cos^(2)B).` Also, ` (tan^(2)A - tan^(2)B)= (sin^(2)A)/(cos^(2)A)- (sin^(2)B)/(cos^(2)B) ` ` = (sin^(2)A cos^(2)B - cos^(2)A sin^(2)B)/(cos^(2)A cos^(2)B) ` ` = ((1- cos^(2)A)cos^(2)B - cos^(2)A(1- cos^(2)B))/(cos^(2)B cos^(2)A) ` ` = ((cos^(2)B - cos^(2)A))/(cos^(2)B cos^(2)A). ` Hence, `(tan^(2)A - tan^(2)B) = ((sin^(2)A - sin^(2)B))/(cos^(2)A cos^(2)B) = ((cos^(2)B - cos^(2)A))/(cos^(2)B cos^(2)A). ` |
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