# Prove that

Question:

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

Solution:

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

$=\left[\frac{(\cot \mathrm{A}+\cos \mathrm{A})-(\cot \mathrm{A}-\cos \mathrm{A})}{(\cot \mathrm{A}+\cos \mathrm{A})+(\cot \mathrm{A}-\cos \mathrm{A})}\right]^{2}+\left[\frac{(\cot \mathrm{A}+\cos \mathrm{A})-(\cot \mathrm{A}-\cos \mathrm{A})}{2}\right]^{2}$

$=\left[\frac{\cot \mathrm{A}+\cos \mathrm{A}-\cot \mathrm{A}+\cos \mathrm{A}}{\cot \mathrm{A}+\cos \mathrm{A}+\cot \mathrm{A}-\cos \mathrm{A}}\right]^{2}+\left[\frac{\cot \mathrm{A}+\cos \mathrm{A}-\cot \mathrm{A}+\cos \mathrm{A}}{2}\right]^{2}$

$=\left[\frac{2 \cos \mathrm{A}}{2 \cot \mathrm{A}}\right]^{2}+\left[\frac{2 \cos \mathrm{A}}{2}\right]^{2}$

$=\left[\frac{\cos \mathrm{A}}{\left(\frac{\cos \mathrm{A}}{\sin \mathrm{A}}\right)}\right]^{2}+[\cos \mathrm{A}]^{2}$

$=\left[\frac{\sin \mathrm{A} \cos \mathrm{A}}{\cos \mathrm{A}}\right]^{2}+[\cos \mathrm{A}]^{2}$

$=[\sin \mathrm{A}]^{2}+[\cos \mathrm{A}]^{2}$

$=\sin ^{2} \mathrm{~A}+\cos ^{2} \mathrm{~A}$

$=1$

$=$ RHS