Question

JEE MAINS SOLVE THE ATTACHED QUESTION ❓

Answer 1

$\bf\dfrac{A}{2}+\dfrac{B}{2}=\dfrac{\pi}{4}~ since~ C=\dfrac{\pi}{2}.$

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$\bf And,~cot\dfrac{A}{2}+cot\dfrac{B}{2}=-\dfrac{q}{p}$

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$\bf\qquad:\implies\:\dfrac{tan\dfrac{A}{2}+tan\dfrac{B}{2}}{tan\dfrac{A}{2}~tan\dfrac{B}{2}}=-\dfrac{q}{p}$

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$\bf And,~cot\dfrac{A}{2}~cot\dfrac{B}{2}=\dfrac{r}{p}$

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$\bf\qquad:\implies~tan\dfrac{A}{2}~tan\dfrac{B}{2}=\dfrac{p}{r}$

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$\bf\qquad:\implies~ tan\dfrac{A}{2}+tan\dfrac{B}{2}=-\dfrac{q}{r}$

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$\bf Then :$

$\bf \qquad:\implies~tan\bigg\lgroup\dfrac{A}{2}+\dfrac{B}{2}\bigg\rgroup=1$

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$\bf\qquad:\implies~ \dfrac{tan\dfrac{A}{2}+tan\dfrac{B}{2}}{1-tan\dfrac{A}{2}~tan\dfrac{B}{2}}=1$

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$\bf \qquad : \implies \: - \dfrac{q}{q} = 1 - \dfrac{p}{r}$

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$\bf \qquad : \implies \: \dfrac{p}{r} = 1 + \dfrac{q}{r}$

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$\qquad \bf : \implies \: \dfrac{p}{r} = \dfrac{r + q}{r}$

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$\qquad \bf : \implies \: p = q + r$

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$\sf\therefore~\underline{The~ correct~ answer~is~\pmb{\sf p=q+r~[OptionA].}}$