Question

Prove that:

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

Answer 1

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

L.H.S.-

$(\csc A-\sin A)(\sec A-\cos A)$

$=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)(\because \csc A=\frac{1}{\sin A}\&\sec A=\frac{1}{\cos A})$

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

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

$=\sin A\cos A.....\left(1\right)$

Now R.H.S.-

$\frac{1}{\tan A+\cot A}$

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

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

$=\sin A\cos A.....\left(2\right)(\because {\sin }^{2}A+{\cos }^{2}A=1)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

L.H.S. $=$ R.H.S.

Hence proved.