Question

Prove that $\frac{\tan A}{\left(secA+1\right)}+\frac{cotA}{\left(\cos ecA+1\right)}=\cos ecA+secA-\cos ecA\times secA$.

Answer 1

STEP 1 : Solving the Left Hand Side (LHS) of the equation

Taking the LHS and solving we get,

$\frac{\tan A}{\left(secA+1\right)}+\frac{cotA}{\left(\cos ecA+1\right)}$

$=\frac{\tan A}{\left(secA+1\right)}\times \frac{\left(secA-1\right)}{\left(secA-1\right)}+\frac{cotA}{\left(\cos ecA+1\right)}\times \frac{\left(\cos ecA-1\right)}{\left(\cos ecA-1\right)}$

$=\frac{\tan A\left(secA-1\right)}{\left({sec}^{2}A-1\right)}+\frac{cotA\left(\cos ecA-1\right)}{\left({\cos ec}^{2}A-1\right)}$

$=\frac{\tan A\left(secA-1\right)}{{\tan }^{2}A}+\frac{cotA\left(\cos ecA-1\right)}{co{t}^{2}A}$

$=\frac{\left(secA-1\right)}{\tan A}+\frac{\left(\cos ecA-1\right)}{cotA}$

$=\frac{secA}{\tan A}-\frac{1}{\tan A}+\frac{\cos ecA}{cotA}-\frac{1}{cotA}$

$=\frac{{\displaystyle \frac{1}{\cos A}}}{{\displaystyle \frac{\sin A}{\cos A}}}-\frac{1}{\tan A}+\frac{{\displaystyle \frac{1}{\sin A}}}{{\displaystyle \frac{\cos A}{\sin A}}}-\frac{1}{cotA}$

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

$=\cos ecA-\frac{\cos A}{\sin A}+secA-\frac{\sin A}{\cos A}$

$=\cos ecA+secA-\frac{\cos A}{\sin A}-\frac{\sin A}{\cos A}$

$=\cos ecA+secA-\left(\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}\right)$

$=\cos ecA+secA-\left(\frac{{\cos }^{2}A+{\sin }^{2}A}{\sin A\times \cos A}\right)$

$=\cos ecA+secA-\left(\frac{1}{\sin A\times \cos A}\right)$

$=\cos ecA+secA-\left(\cos ecA\times secA\right)$

i.e. $\frac{\tan A}{\left(secA+1\right)}+\frac{cotA}{\left(\cos ecA+1\right)}=\cos ecA+secA-\cos ecA\times secA$

Hence proved.