Question

Prove that $\frac{(\cos A-\sin A+1)}{(\cos A+\sin A-1)}=\cos ecA+cotA.$

Answer 1

$L.H.S=\frac{(\cos A-\sin A+1)}{(\cos A+\sin A-1)}$

Divide numerator and denominator with $\sin A$.

$=\frac{({\displaystyle \frac{\cos A}{\sin A}}-{\displaystyle \frac{\sin A}{\sin A}}+{\displaystyle \frac{1}{\sin A}})}{({\displaystyle \frac{\cos A}{\sin A}}+{\displaystyle \frac{\sin A}{\sin A}}-{\displaystyle \frac{1}{\sin A}})}$

$=\frac{(cotA–1+\cos ecA}{(cotA+1–\cos ecA)}$

$=\frac{\left(cotA+\cos ecA\right)-\left(\cos e{c}^{2}A-co{t}^{2}A\right)}{(cotA+1–\cos ecA)}$……………………..$\left[\left(1+co{t}^2\theta =cose{c}^2\theta \right)\right]$

$=\frac{\left(cotA+\cos ecA\right)-\left(\cos ecA+cotA\right)\left(\cos ecA-cotA\right)}{\left(cotA+1-\cos ecA\right)}$………………….($\mathbf{[}{\mathit{a}}^{\mathbf{2}}\mathbf{-}{\mathit{b}}^{\mathbf{2}}\mathbf{=}\left(a+b\right)\left(a-b\right)\mathbf{}\mathbf{]}$)

$=\frac{(cotA+\cos ecA)(1–\cos ecA+cotA)}{(cotA+1–\cos ecA)}$

$=\frac{(cotA+\cos ecA)(1–\cos ecA+cotA)}{(1–\cos ecA+cotA)}$

$=(cotA+\cos ecA)$

$=R.H.S.$

Hence, it is proved that, $\frac{(\cos A-\sin A+1)}{(\cos A+\sin A-1)}=\cos ecA+cotA.$