Question

if $x=\left(\cot A+\cos A\right)\text{and}y=\left(\cot A-\cos A\right)$ prove that ${\left(\frac{x-y}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2=1$

Answer 1

Given,

$x=\cot A+\cos A$

$y=\cot A-\cos A$

${\left(\frac{x-y}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2$

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

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

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

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

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

$=1$