If x = (sec A + sin A) and y = (sec A – sin A), prove that


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




x = secA + sinA       .....(1)

= secA – sinA       .....(2)

Adding (1) and (2), we get

$x+y=\sec A+\sin A+\sec A-\sin A$

$\Rightarrow 2 \sec A=x+y$

$\Rightarrow \sec A=\frac{x+y}{2}$


$\Rightarrow \frac{1}{\sec A}=\frac{2}{x+y}$

$\Rightarrow \cos A=\frac{2}{x+y} \quad \ldots \ldots(3)$

Subtracting (2) from (1), we get

$x-y=\sec A+\sin A-\sec A+\sin A$

$\Rightarrow 2 \sin A=x-y$

$\Rightarrow \sin A=\frac{x-y}{2} \quad \ldots \ldots(4)$

We know

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

$\Rightarrow\left(\frac{2}{x+y}\right)^{2}+\left(\frac{x-y}{2}\right)^{2}=1 \quad[$ Using $(3)$ and $(4)]$



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