If $\frac{cotθ+cosθ}{cotθ-cosθ}=\frac{k+1}{1-k}, k ≠ 1,$ then k is equal to :

sinθ

\(\frac{cotθ + cosθ}{cotθ - cosθ}\) = \(\frac{k + 1 }{1 - k }\)

By applying componendo and dividendo,

(\frac{cotθ + cosθ + cotθ - cosθ}{cotθ + cosθ - cotθ +cosθ}\) = \(\frac{k + 1 +1 - k }{k + 1 -1 + k }\)

(\frac{ cotθ }{cosθ }\) = \(\frac{1 }{k }\)

cosec θ  = \(\frac{1 }{k }\)

sin θ = k

Ans :- sin θ