Question

Proove:

$(cosecA-sinA)(secA-cosA)=\frac{1}{tanA+cotA}$

Answer 1

$(cosecA-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$

L.H.S :- $(cosecA-\sin A)(\sec A-\cos A)$

$=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)$ $[\because cosecA=\frac{1}{\sin A},\sec A=\frac{1}{\cos A}]$

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

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

$=[\frac{\cos A}{\sin A}.\cos A][\frac{\sin A}{\cos A}.\sin A]$ $[\because \tan A=\frac{\sin A}{\cos A},\cot A=\frac{\cos A}{\sin A}]$

$=\cot A.\cos A.\tan A.\sin A$ $[\tan A.\cot A=1]$

$=\cot A.\tan A.\cos A.\sin A$

$=\cos A.\sin A=$ L.H.S.

R.H.S. $=\frac{1}{\tan A+\cot A}=\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}$

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

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

$=\cos A.\sin A=R.H.S.$

$\therefore$ $L.H.S.=R.H.S$