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Question

If 1+cosθ1cosθ=cosec θ+cotθ, where θ=kπ8,kN, then the number of possible value(s) of θ[0,2π] is

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Solution

Given
1+cosθ1cosθ=cosec θ+cotθ
So,
θ{0,π,2π}
Now,
1+cosθ1cosθ×1+cosθ1+cosθ=cosec θ+cotθ(1+cosθ)21cos2θ=cosec θ+cotθ|1+cosθ||sinθ|=cosec θ+cotθ1+cosθ|sinθ|=cosec θ+cotθ (1+cosθ0)1+cosθ|sinθ|=1+cosθsinθ
It is possible if
sinθ>0π>θ>0π>kπ8>0k(0,8)

Hence, there are 7 possible values of k.

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