Question

If $x=\text{cosec}A+\cos A$ and $y=\text{cosec}A-\cos A$ then prove that ${\left(\frac{2}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2-1=0$.

Answer 1

Given:-

$\Rightarrow$$x=cosecA+\cos A$

$\Rightarrow$$y=cosecA-\cos A$

To prove:-

$\Rightarrow$${\left(\frac{2}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2-1=0$

$\therefore x+y=cosecA+\cos A+cosecA-\cos A=2cosecA$

$x-y=cosecA+\cos A-cosecA+\cos A=2\cos A$

$\therefore {\left(\frac{2}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2-1={\left(\frac{2}{2cosecA}\right)}^2+{\left(\frac{2\cos A}{2}\right)}^2-1$

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

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

$\Rightarrow 1-1$

$\Rightarrow 0$

Hence proved.