Question

\(\frac{1+c\text{os}x-s\text{inx}}{1-c\text{os}x-s\text{inx}}=-cot\frac{x}{2}\)

Answer 1

\(\frac{1+cosx-sinx}{1-cosx-sinx}=\frac{1+2cos^2\frac{x}{2}-1-2sin\frac{x}{2}.cos\frac{x}{2}}{1-1+2sin^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}=\frac{2cos^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}{2sin^2\frac{x}{2}-2sin\frac{x}{2}.cos\frac{x}{2}}\)

\(=\frac{-2cos\frac{x}{2}\left(sin\frac{x}{2}-cos\frac{x}{2}\right)}{2sin\frac{x}{2}\left(sin\frac{x}{2}-cos\frac{x}{2}\right)}=\frac{-cos\frac{x}{2}}{sin\frac{x}{2}}=-cot\frac{x}{2}\)