Question

Proove that $\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}=\cos A+\sin A$

Answer 1

Given $\frac{cosA}{1-tanA}+\frac{sinA}{1-cotA}$ can be simplified as

$\frac{cosA}{1-\frac{sinA}{cosA}}+\frac{sinA}{1-\frac{cosA}{sinA}}$ -----$(tanA=\frac{sinA}{cosA}$ and $cotA=\frac{cosA}{sinA}$ $)$

=$\frac{co{s}^{2}A}{cosA-sinA}+\frac{si{n}^{2}A}{sinA-cosA}$

=$\frac{co{s}^{2}A}{cosA-sinA}-\frac{si{n}^{2}A}{cosA-sinA}$

=$\frac{co{s}^{2}A-si{n}^{2}A}{cosA-sinA}$

=$\frac{(cosA-sinA)(cosA+sinA)}{(cosA-sinA)}$

=$cosA+sinA$