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Question Number 52609 by subhankar10 last updated on 10/Jan/19

sin^2 A + sin^4 A=1then the value of tan^2 A−tan^4 A

sin2A+sin4A=1thenthevalueoftan2A−tan4A

Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19

sin^4 A=1−sin^2 A sin^2 A×sin^2 A=cos^2 A tan^2 A=cosec^2 A tan^2 A−tan^4 A tan^2 A(1−tan^2 A) =cosec^2 A(1−cosec^2 A) =(1/(sin^2 A))(((sin^2 A−1)/(sin^2 A))) =((sin^2 A−sin^2 A−sin^4 A)/(sin^4 A))[since sin^2 A+sin^4 A=1] =−1

sin4A=1−sin2Asin2A×sin2A=cos2Atan2A=cosec2Atan2A−tan4Atan2A(1−tan2A)=cosec2A(1−cosec2A)=1sin2A(sin2A−1sin2A)=sin2A−sin2A−sin4Asin4A[sincesin2A+sin4A=1]=−1

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