Question

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

Answer 1

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

$L.H.S.=(\cos ecA-\sin A)(secA-\cos A)$

$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)$

$=\left(\frac{1-{\sin }^{2}A}{\sin A}\right)\left(\frac{1-{\cos }^{2}A}{\cos A}\right)$

$=\left(\frac{{\cos }^{2}A}{\sin A}\right)\left(\frac{{\sin }^{2}A}{\cos A}\right)$ ……….($1-{\cos }^{2}A={\sin }^{2}A,{\sin }^{2}A+{\cos }^{2}A=1$)

$=\frac{{\cos }^{2}A.{\sin }^{2}A}{\sin A.\cos A}$

$=\frac{\cos A.\sin A}{1}$

$=\frac{\sin A.\cos A}{{\sin }^{2}A+{\cos }^{2}A}$ ..……(${\sin }^{2}A+{\cos }^{2}A=1$)

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

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

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

$=\frac{1}{\tan A+cotA}$

$=R.H.S.$

Hence proved.