Given
√1+cosθ1−cosθ=cosec θ+cotθ
So,
θ≠{0,π,2π}
Now,
√1+cosθ1−cosθ×1+cosθ1+cosθ=cosec θ+cotθ⇒√(1+cosθ)21−cos2θ=cosec θ+cotθ⇒|1+cosθ||sinθ|=cosec θ+cotθ⇒1+cosθ|sinθ|=cosec θ+cotθ (∵1+cosθ≥0)⇒1+cosθ|sinθ|=1+cosθsinθ
It is possible if
sinθ>0⇒π>θ>0⇒π>kπ8>0∴k∈(0,8)
Hence, there are 7 possible values of k.