Question

If x = (sec A + sin A) and y = (sec A – sin A), prove that ${\left(\frac{2}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2=1$.

Answer 1

Given:

x = secA + sinA .....(1)

y = secA – sinA .....(2)

Adding (1) and (2), we get

$$\begin{gathered} x+y=\sec A+\sin A+\sec A-\sin A \\ \Rightarrow 2\sec A=x+y \\ \Rightarrow \sec A=\frac{x+y}{2} \\ \Rightarrow \frac{1}{\sec A}=\frac{2}{x+y} \end{gathered}$$
$\Rightarrow \cos A=\frac{2}{x+y}.....\left(3\right)$

Subtracting (2) from (1), we get

$$\begin{gathered} x-y=\sec A+\sin A-\sec A+\sin A \\ \Rightarrow 2\sin A=x-y \\ \Rightarrow \sin A=\frac{x-y}{2}.....\left(4\right) \end{gathered}$$

We know

$$\begin{gathered} {\cos }^{2}A+{\sin }^{2}A=1 \\ \Rightarrow {\left(\frac{2}{x+y}\right)}^2+{\left(\frac{x-y}{2}\right)}^2=1\left[\mathrm{Using}\left(3\right)\mathrm{and}\left(4\right)\right] \end{gathered}$$