Question

if (a^2-b^2)sintheta +2ab costheta =a^2+b^2, then prove that tantheta =a^2-b^2/2ab

Answer 1

(a2−b2)sinθ+2abcosθ=a2+b2

⇒a2−b2a2+b2sinθ+2aba2+b2cosθ=1.....(1)

Now if a2−b2a2+b2=cosα

then

sinα=√1−cos2α

=√1−(a2−b2a2+b2)2

=2aba2+b2

So the relation (1) becomes

cosαsinθ+sinαcosθ=1

⇒sin(θ+α)=sin(π2)

⇒α=π2−θ

⇒cosα=cos(π2−θ)

⇒cosα=sinθ

∴sinθ=cosα=a2−b2a2+b2

Alternative-1

Given

(a2−b2)sinθ+2abcosθ=a2+b2

Transferring to RHS we get

a2(1−sinθ)+b2(1+sinθ)−2abcosθ=0

⇒(a√1−sinθ)2+(b√1+sinθ)2−2ab√1−sin2θ=0

⇒[(a√1−sinθ)−(b√1+sinθ)]2=0

⇒a√1−sinθ=b√1+sinθ

⇒a2(1−sinθ)=b2(1+sinθ)

⇒sinθ=a2−b2