In a ,△PQR,R_=π2 If tan P2 and tan Q2 are the roots of ax2+bx+c=0,a≠0 , then

Since R=π2⇒P_+Q=π2⇒P_2=π4-QQ2∴tan P2=tan π4-Q2=tan π4-tan Q2a+tan π4tan Q2=tan P2+tan P2tan Q2=1-tan Q2=tan P2+tan Q2=1-tan P2tan Q2⇒-ba=1-ca⇒-b=a-c⇒c=a+b

In a ,△PQR,R_=π2 If tan P2 and tan Q2 are the roots of ax2+bx+c=0,a≠0 , then

Since R=π2⇒P_+Q=π2⇒P_2=π4-QQ2∴tan P2=tan π4-Q2=tan π4-tan Q2a+tan π4tan Q2=tan P2+tan P2tan Q2=1-tan Q2=tan P2+tan Q2=1-tan P2tan Q2⇒-ba=1-ca⇒-b=a-c⇒c=a+b

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In a , $△ PQR, \underline{R} = \frac{π}{2}$ If $\tan (\frac{P}{2})$ and $\tan (\frac{Q}{2})$ are the roots of ${ax}^{2} + bx + c = 0, a \neq 0$ , then

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$b = a + c$

$b = c$

$c = a + b$

$a = b + c$

answer is C.

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Detailed Solution

Since $⌊R = \frac{π}{2} \Rightarrow \underline{P} + Q = \frac{π}{2} \Rightarrow \frac{\underline{P}}{2} = \frac{π}{4} - \frac{QQ}{2} ∴ \tan \frac{P}{2} = \tan (\frac{π}{4} - \frac{Q}{2})$ $= \frac{\tan \frac{π}{4} - \tan \frac{Q}{2}}{a + \tan \frac{π}{4} \tan \frac{Q}{2}} = \tan \frac{P}{2} + \tan \frac{P}{2} \tan \frac{Q}{2} = 1 - \tan \frac{Q}{2} = \tan \frac{P}{2} + \tan \frac{Q}{2} = 1 - \tan \frac{P}{2} \tan \frac{Q}{2}$ $\Rightarrow - \frac{b}{a} = 1 - \frac{c}{a} \Rightarrow - b = a - c \Rightarrow c = a + b$