Question

if tanA=b/a then show that a cos 2A+ b sin2A= a​

Answer 1

Answer:

Given that tan A = b/a, prove that a cos 2A + b sin 2A = a.

So sin A = b/(a^2+b^2)^0.5 and cos A = a/(a^2+b^2)^0.5

a cos 2A = a[cos^2 A - sin^2 A] = a[a^2-b^2]/(a^2+b^2) …(1)

b sin 2A = b[2sin A cos A] = 2b[ab]/(a^2+b^2) …(2)

a cos 2A + b sin 2A = sum of (1) and (2)

= a[a^2-b^2]/(a^2+b^2) + 2b[ab]/(a^2+b^2)

= [a^3-ab^2+2ab^2]/(a^2+b^2)

= a[a^2+b^2]/(a^2+b^2]

= a. Proved

Answer 2

Answer:

Given that tan A = b/a, prove that a cos 2A + b sin 2A = a.

So sin A = b/(a^2+b^2)^0.5 and cos A = a/(a^2+b^2)^0.5

a cos 2A = a[cos^2 A - sin^2 A] = a[a^2-b^2]/(a^2+b^2) …(1)

b sin 2A = b[2sin A cos A] = 2b[ab]/(a^2+b^2) …(2)

a cos 2A + b sin 2A = sum of (1) and (2)

= a[a^2-b^2]/(a^2+b^2) + 2b[ab]/(a^2+b^2)

= [a^3-ab^2+2ab^2]/(a^2+b^2)

= a[a^2+b^2]/(a^2+b^2]

= a. Proved

Step-by-step explanation:

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