Hi,
Please help.
Prove that:
(sin θ-kos θ+1)/(sin θ+kos θ-1)= (sin θ+1)/(kos θ)
(kos θ/ 1-tan x) + (sin θ/1-kot θ) = sin θ + kos θ
Solve that tanx = π - 2x , 0<x<π/2
Thnks.
1) $\displaystyle \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2} +2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}-2\sin^2\frac{\theta}{2}}=\frac{\sin\frac{\theta}{2 }+\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}=\\=\frac{\left(\sin \frac{\theta}{2}+\cos \frac{\theta}{2}\right)^2}{\cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2}}=\frac{1+\sin \theta}{\cos \theta}$
2) $\displaystyle \frac{\cos \theta}{1-\tan \theta}+\frac{\sin \theta}{1-\cot \theta}=\frac{\cos^2 \theta}{\cos \theta-\sin \theta}+\frac{\sin^2 \theta}{\sin \theta-\cos \theta}=\\=\frac{\cos^2 \theta-\sin^2 \theta}{\cos \theta-\sin \theta}=\cos \theta+\sin \theta$
I don't understand the last part of the problem