If $x=\operatorname{cosec} \mathrm{A}+\cos \mathrm{A}$ and $y=\operatorname{cosec} \mathrm{A}-\cos \mathrm{A}$, then prove that $\left(\frac{2}{x+y}\right)^{2}+\left(\frac{x-y}{2}\right)^{2}-1=0$
LHS $=\left(\frac{2}{x+y}\right)^{2}+\left(\frac{x-y}{2}\right)^{2}-1$
$=\left[\frac{2}{(\operatorname{cosec} A+\cos A)+(\operatorname{cosec} A-\cos A)}\right]^{2}+\left[\frac{(\operatorname{cosec} A+\cos A)-(\operatorname{cosec} A-\cos A)}{2}\right]^{2}-1$
$=\left[\frac{2}{\operatorname{cosec} A+\cos A+\operatorname{cosec} A-\cos A}\right]^{2}+\left[\frac{\operatorname{cosec} A+\cos A-\operatorname{cosec} A+\cos A}{2}\right]^{2}-1$
$=\left[\frac{2}{2 \operatorname{cosec} A}\right]^{2}+\left[\frac{2 \cos A}{2}\right]^{2}-1$
$=\left[\frac{1}{\operatorname{cosec} A}\right]^{2}+[\cos A]^{2}-1$
$=[\sin \mathrm{A}]^{2}+[\cos \mathrm{A}]^{2}-1$
$=\sin ^{2} \mathrm{~A}+\cos ^{2} \mathrm{~A}-1$
$=1-1$
$=0$
$=\mathrm{RHS}$
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