Question

If $\cos A-\sin A=mand\cos A+\sin A=n$, show that $\frac{m^2-n^2}{m^2+n^2}=-2\sin A\cos A=-\frac{2}{\tan A+\cot A}$

Answer 1

Given $cosA-sinA=m$ and $cosA+sinA=n$

want to show : $\frac{m^2-n^2}{m^2+n^2}=-2sinAcosA=\frac{-2}{tanA+cotA}$

proof :- firstly observe that $\frac{-2}{tanA+cotA}=\frac{-2}{\frac{sinA}{cosA}+\frac{cosA}{sinA}}=\frac{2sinAcosA}{si{n}^{2}A+co{s}^{2}A}$

$=-2sinAcosA$

$m^2-n^2=(cosA-sinA{)}^2-(cosA+sinA{)}^2$

$=(co{s}^{2}A+si{n}^{2}A-2cosAsinA)-(co{s}^{2}A+si{n}^{2}A=2cosAsinA)$

$=-4cosAsinA$

$m^2-n^2=(cosA-sinA{)}^2-(cosA+sinA{)}^2$

$=(co{s}^{2}A+si{n}^{2}A-2cosAsinA)-(co{s}^{2}A+si{n}^{2}A=2cosAsinA)$

$=2(co{s}^{2}A+si{n}^{2}A)$

$=2$

thus , $\frac{m^2-n^2}{m^2+n^2}=-2cosAsinA=\frac{-2}{tanA+cotA}$