Question

if tanA=b/a then show that acos2A+bsin2A=a​

Answer 1

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Given,

tan A = b/a

To Show,

acos2A + bsin2A = a

LHS = acos2A + bsin2A

= (1-tan²A)a / 1+tan²A + 2(tanA)b / 1+tan²A

= a(1-tan²A)+2(tanA)b / 1+tan²A

= a(1-b²/a²)+2b²/a²

= a³-ab²+2ab² / a²+b²

= a³+ab² / a²+b²

= a(a²+b²) / (a²+b²)

= a = RHS.

LHS = RHS

Hence Proved.