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