Question
If $\text{x}=\text{cosec}\text{A}+\cos\text{A}$ and $\text{y}=\text{cosec}\text{A}-\cos\text{A}$ then Prove that $\Big(\frac{2}{\text{x}+\text{y}}\Big)^2+\Big(\frac{\text{x}-\text{y}}{2}\Big)^2-1=0.$
✓
Answer
$\text{x}=\text{cosec}\text{A}+\cos\text{A}$ and $\text{y}=\text{cosec A}-\cos\text{A}$
Thus, we have
$\text{x}+\text{y}=(\text{cosec A}+\cos\text{A})\\+(\text{cosec A}-\cos\text{A})=2\text{cosec A}$
$\text{x}-\text{y}=(\text{cosec A}+\cos\text{A})\\-(\text{cosec A}-\cos\text{A})=2\cos\text{A}$
$\text{L.H.S.}=\Big(\frac{2}{\text{x}+\text{y}}\Big)^2+\Big(\frac{\text{x}-\text{y}}{2}\Big)^2-1$
$=\Big(\frac{2}{2\text{cosec A}}\Big)^2+\Big(\frac{2\cos\text{A}}{2}\Big)^2-1$
$=\Big(\frac{1}{\text{cosec A}}\Big)^2+(\cos\text{A})^2-1$
$=(\sin\text{A})^2+(\cos\text{A})^2-1$
$=\sin^2\text{A}+\cos^2\text{A}-1$
$=1-1$
$=0$
$=\text{R.H.S}$
Thus, we have
$\text{x}+\text{y}=(\text{cosec A}+\cos\text{A})\\+(\text{cosec A}-\cos\text{A})=2\text{cosec A}$
$\text{x}-\text{y}=(\text{cosec A}+\cos\text{A})\\-(\text{cosec A}-\cos\text{A})=2\cos\text{A}$
$\text{L.H.S.}=\Big(\frac{2}{\text{x}+\text{y}}\Big)^2+\Big(\frac{\text{x}-\text{y}}{2}\Big)^2-1$
$=\Big(\frac{2}{2\text{cosec A}}\Big)^2+\Big(\frac{2\cos\text{A}}{2}\Big)^2-1$
$=\Big(\frac{1}{\text{cosec A}}\Big)^2+(\cos\text{A})^2-1$
$=(\sin\text{A})^2+(\cos\text{A})^2-1$
$=\sin^2\text{A}+\cos^2\text{A}-1$
$=1-1$
$=0$
$=\text{R.H.S}$
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