Question

Prove that : ${(1-\tan A)}^2+{(1-\cot A)}^2={(\sec A-\csc A)}^2$

Answer 1

L.H.S$={(1-\tan A)}^2+{(1-\cot A)}^2$

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

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

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

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

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

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

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

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

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

$={(\sec A-\csc A)}^2=$R.H.S

Hence proved.