[JEE Advanced 1999] In $\triangle PQR$, $\angle R=\frac\pi2$. If $\tan\frac P2,\tan\frac Q2$ are the roots of $ax^2+bx+c=0$ ($a\ne0$), then
In $\triangle PQR$, $\angle R=\frac\pi2$. If $\tan\frac P2,\tan\frac Q2$ are the roots of $ax^2+bx+c=0$ ($a\ne0$), then
(a) $a+b=c$
(b) $b+c=a$
(c) $a+c=b$
(d) $b=c$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
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Correct answer: (a) $a+b=c$
$\frac P2+\frac Q2=\frac\pi4$, so $\tan\frac P2+\tan\frac Q2=1-\tan\frac P2\tan\frac Q2$, i.e. $-\frac ba=1-\frac ca\Rightarrow a+b=c$.
JEE Advanced 1999 · Trigonometry — verified solution by the Vidaara Team.
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