Question

$\frac{\cos A}{1-\tan A}-\frac{{\sin }^{2}A}{\cos A-\sin A}=\sin A+\cos A$.

Answer 1

We have,

L.H.S. = $\frac{cosA}{1-tanA}+\frac{si{n}^{2}A}{sinA-cosA}$

$\Rightarrow$ L.H.S. = $\frac{cosA}{1-\frac{sinA}{cosA}}+\frac{si{n}^{2}A}{sinA-cosA}$

$\Rightarrow$ L.H.S. = $\frac{cosA}{\frac{cosA-sinA}{cosA}}+\frac{si{n}^{2}A}{sinA-cosA}$

$\Rightarrow$ L.H.S. = $\frac{co{s}^{2}A}{cosA-sinA}+\frac{si{n}^{2}A}{sinA-cosA}$

$\Rightarrow$ L.H.S. = $\frac{co{s}^{2}A}{cosA-sinA}-\frac{si{n}^{2}A}{cosA-sinA}$

$\Rightarrow$ L.H.S. = $\frac{co{s}^{2}A-si{n}^{2}A}{cosA-sinA}$

$\Rightarrow$ L.H.S. = $\frac{(cosA+sinA)(cosA-sinA)}{cosA-sinA}=cosA+sinA=R.H.S.$