In triangle PQR, ∠R = π/2, if tan P/2 and tan Q/2 are roots of quadratic equation ax²+bx+c=0 where a ≠ 0 then (a) b=c (b) b = a+c (c) a= b+c (d) c = a+b

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Arun · Answer

ax^2 +bx + c =0
sum of roots tanP/2 + tanQ/2 = -b/a
product of roots tanP/2 tanQ/2 = c/a
tan (P/2 +Q/2 ) =( tanP/2 + tanQ/2 )/1- tanP/2 tanQ/2
or tan 45 = {-b/a }/ 1- c/a [ P +Q = 90 so P/2 + Q/2 = 45 ]
or 1 = -b / (a-c)
or a - c = -b
or a + b -c =0 Q . E. D.

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In triangle PQR, ∠R = π/2, if tan P/2 and tan Q/2 are roots of quadratic equation ax²+bx+c=0 where a ≠ 0 then (a) b=c (b) b = a+c (c) a= b+c (d) c = a+b

In triangle PQR, ∠R = π/2, if tan P/2 and tan Q/2 are roots of quadratic equation ax²+bx+c=0 where a ≠ 0 then
(a) b=c
(b) b = a+c
(c) a= b+c
(d) c = a+b

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In triangle PQR, ∠R = π/2, if tan P/2 and tan Q/2 are roots of quadratic equation ax²+bx+c=0 where a ≠ 0 then (a) b=c (b) b = a+c (c) a= b+c (d) c = a+b

In triangle PQR, ∠R = π/2, if tan P/2 and tan Q/2 are roots of quadratic equation ax²+bx+c=0 where a ≠ 0 then(a) b=c(b) b = a+c(c) a= b+c(d) c = a+b

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In triangle PQR, ∠R = π/2, if tan P/2 and tan Q/2 are roots of quadratic equation ax²+bx+c=0 where a ≠ 0 then
(a) b=c
(b) b = a+c
(c) a= b+c
(d) c = a+b