Question

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Answer 1

$\large\underline{\sf{Solution-}}$

Given that ,

In triangle ABC,

$\rm :\longmapsto\:\dfrac{ {sin}^{2}A + {sin}^{2}B + {sin}^{2}C}{ {cos}^{2}A + {cos}^{2}B + {cos}^{2}C} = 2$

We know,

$\: \: \: \: \: \: \: \: \: \: \: \red{\bigg \{\: {sin}^{2}x + {cos}^{2}x = 1 \bigg \}}$

Using this identity, in numerator we get

$\rm :\longmapsto\:\dfrac{ 1 - {cos}^{2}A +1 - {cos}^{2}B + 1 - {cos}^{2}C}{ {cos}^{2}A + {cos}^{2}B + {cos}^{2}C} = 2$

$\rm :\longmapsto\:\dfrac{ 3 - {cos}^{2}A- {cos}^{2}B- {cos}^{2}C}{ {cos}^{2}A + {cos}^{2}B + {cos}^{2}C} = 2$

$\rm :\longmapsto\:3 - ( {cos}^{2}A + {cos}^{2}B + {cos}^{2}C) = 2({cos}^{2}A + {cos}^{2}B + {cos}^{2}C)$

$\rm :\longmapsto\:3 = 3({cos}^{2}A + {cos}^{2}B + {cos}^{2}C)$

$\rm :\longmapsto\:{cos}^{2}A + {cos}^{2}B + {cos}^{2}C = 1$

$\rm :\longmapsto\:{cos}^{2}A + {cos}^{2}B = 1 - {cos}^{2}C$

$\rm :\longmapsto\:{cos}^{2}A + {cos}^{2}B = {sin}^{2}C$

$\rm :\longmapsto\:{cos}^{2}A + {cos}^{2}B - {sin}^{2}C = 0$

$\rm :\longmapsto\:{cos}^{2}A + cos(B + C)cos(B - C) = 0$

$\red{\bigg \{ \because \: {cos}^{2}x - {sin}^{2}y = cos(x + y)cos(x - y) \bigg \}}$

$\rm :\longmapsto\:{cos}^{2}A + cos(\pi - A)cos(B - C) = 0$

$\red{\bigg \{ \because \: A + B + C = \pi\bigg \}}$

$\rm :\longmapsto\:{cos}^{2}A - cos(A)cos(B - C) = 0$

$\red{\bigg \{ \because \: cos(\pi - x) = - cosx\bigg \}}$

$\rm :\longmapsto\:cosA{\bigg(cosA - cos(B - C) \bigg) } = 0$

$\rm :\implies\:cosA = 0 \: \: \: or \: \: \: cosA = cos(B - C)$

Case :- 1

$\rm :\longmapsto\:cosA = 0$

$\bf\implies \:A = \dfrac{\pi}{2}$

$\bf\implies \: \triangle \: is \: right \: angled$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{ \: Option \: (a) \: is \: correct}}}$

Case :- 2

$\rm :\longmapsto\:cosA = cos(B - C)$

$\rm :\longmapsto\:A = B - C$

$\rm :\longmapsto\:B + B= B + C + A$

$\rm :\longmapsto\:2B= \pi$

$\bf\implies \:B = \dfrac{\pi}{2}$

$\bf\implies \: \triangle \: is \: right \: angled$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf{ \: Option \: (a) \: is \: correct}}}$