Question

Prove that $\frac{(\tan A+\sin A)}{(\tan A-\sin A)}=\frac{(secA+1)}{(secA-1)}$.

Answer 1

Solving the left hand side (LHS) of the equation:

Taking the LHS and solving we get,

$\frac{\left(\tan A+\sin A\right)}{\left(\tan A-\sin A\right)}$

$=\frac{\left({\displaystyle \frac{\sin A}{\cos A}}+\sin A\right)}{\left({\displaystyle \frac{\sin A}{\cos A}}-\sin A\right)}$ $\left\{\tan A=\frac{\sin A}{\cos A}\right\}$

$=\frac{\left[\sin A\left({\displaystyle \frac{1}{\cos A}+1}\right)\right]}{\left[\sin A\left({\displaystyle \frac{1}{\cos A}-1}\right)\right]}$ $\left\{secA=\frac{1}{\cos A}\right\}$

$=\frac{\left({\displaystyle secA+1}\right)}{\left({\displaystyle secA-1}\right)}=RHS$

i.e. $\frac{(\tan A+\sin A)}{(\tan A-\sin A)}=\frac{(secA+1)}{(secA-1)}$

Hence, proved.