To prove: \(\frac{1\,-\,cos2x\,+\,sinx}{sin2x\,+\,cosx}=tanx\)
Taking LHS,
We know that,
1 – cos 2x = 2 sin2x & sin 2x = 2 sinx cosx
Taking sinx common from the numerator and cosx from the denominator
= \(\frac{sinx}{cosx}\)
= tanx [∵ tanθ = \(\frac{sin\theta}{cos\theta}\)]
= RHS
∴ LHS = RHS
Hence Proved