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Question

(a2-b2)sinθ+2abcosθ=a2+b2, find sinθ


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Solution

Step 1: Simplify the given equation.

We are given, (a2-b2)sinθ+2abcosθ=a2+b2,

Dividing by a2+b2 throughout the equation we get,

(a2-b2)sinθa2+b2+2abcosθa2+b2=1...(i)

Say a2-b2a2+b2=cosα...(ii)

Then sinα=1-cos2α

=1-a2-b2a2+b22

=4a2b2a2+b22

sinα=2aba2+b2...(iii)

Step 2: Solve further to get the required answer.

Now, substitute value of cosα and sinα in the equation (i), we get,

cosαsinθ+sinαcosθ=1 ...[sinA+B=sinAcosB+cosAsinB]

sin(α+θ)=1

sin(α+θ)=sinπ2 ...[sinπ2=1]

α+θ=π2

θ=π2-α

sinθ=sinπ2-α

sinθ=cosα ...[sinπ2-A=cosA]

sinθ=a2-b2a2+b2 …[From (ii)]

Hence, the required answer is sinθ=a2-b2a2+b2


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