A triangle `PQR , /_R=90^@` and `tan(P/2)` and `tan(Q/2)` roots of the `ax^2+bx+c=0` then prove that `a+b=c`A. a+b = cB. b+c = 0C. a+c = bD. b = c

A triangle `PQR , /_R=90^@` and `tan(P/2)` and `tan(Q/2)` roots of the

1.	A triangle `PQR , /_R=90^@` and `tan(P/2)` and `tan(Q/2)` roots of the `ax^2+bx+c=0` then prove that `a+b=c`A. a+b = cB. b+c = 0C. a+c = bD. b = c
Answer» Correct Answer - A It is given that `"tan"(P)/(2) and "tan"(Q)/(2)` are roots of `ax^(2) + bc + c = 0`. `therefore" ""tan"(P)/(2) + "tan"(Q)/(2) = - (b)/(a) and "tan"(P)/(2) "tan"(Q)/(2) = (c)/(a)` We have, `R = (pi)/(2) and P+Q+R = pi rArr P+Q = (pi)/(2) rArr (P)/(2)+(Q)/(2) = (pi)/(4)` `therefore" "tan((P)/(2)+(Q)/(2)) = "tan"(pi)/(4)` `rArr" "(tan P//2 + tan Q//2)/(1-tan P//2 tan Q//2) = 1` `rArr" "(-(b)/(a))/(1-(c)/(a))=1` `rArr" "1-(c)/(a) = - (b)/(a) rArr a-c = - b rArr a+b = c`

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A triangle `PQR , /_R=90^@` and `tan(P/2)` and `tan(Q/2)` roots of the `ax^2+bx+c=0` then prove that `a+b=c`A. a+b = cB. b+c = 0C. a+c = bD. b = c

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