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Question

Find the maximum value of 1111 1+sin θ1111+cos θ

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Solution

Let =1111 1+sinθ1111+cosθ

Applying R2R2-R1 and R3R3-R1, we get

=1110 sinθ000cosθ

=sinθcosθ=sin2θ2
We know that −1 ≤ sin2θ ≤ 1.
∴ Maximum value of ∆ = 12×1=12

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