Question

In triangle pqr, angle r=90 degree if tan(p/2) and tan(q/2) are the roots of the equation

Answer 1

Answer:

c = a + b

Step-by-step explanation:

Complete Question : In triangle pqr, angle r=90 degree if tan(p/2) and tan(q/2) are the roots of the equation ax² + bx + c then show that c = a + b

tan(p/2) and tan(q/2) are the roots of the equation  ax² + bx + c

=> tan(p/2) +  tan(q/2) = -b/a  ( sum of roots)

   tan(p/2) *  tan(q/2)  = c/a    (products of roots)

Using Tan(A + B) = (TanA + TanB)/(1 - TanA.TanB)

Tan (p/2 + q/2) = ( tan(p/2) +  tan(q/2) )/(1 - tan(p/2) *  tan(q/2)

In triangle pqr, angle r=90° => p + q = 90°

=> p/2 + q/2 = 45°

=> Tan (45°) = ( tan(p/2) +  tan(q/2) )/(1 - tan(p/2) *  tan(q/2)

=> 1  =  ( tan(p/2) +  tan(q/2) )/(1 - tan(p/2) *  tan(q/2)

=> tan(p/2) +  tan(q/2)  = 1 - tan(p/2) *  tan(q/2

=> -b/a  = 1 - c/a

=> - b = a - c

=> c = a + b

Answer 2

Answer:

use tan A+B formula

tan( p/2+q/2)=tan p/2 +tan q/2 /1-tanp/2×tan q/2

we know that r=90

so p+q+90=180 ( angle sum property)

p+q=90

we get p/2+q/2=45

tan 45 =1

so , 1= tan p/2+tan q/2/1-tan p/2×tan q/2

tan p/2 + tan q/2 = 1- tan p/2 × tan q/2

tan p/2×tan q/2 = 1- tan p/2 +tan q/2

we know that sum of roots = -b/a

product of roots = c/a

so sum = tan p/2 + tan q/2 =-b/a

product = tan p/2 ×tan q/2 = c/a

so we get

c/a = 1-(-b/a)

c/a = 1+ b/a

c/a = (a+b)/a ( MULTIPLYING BOTH SIDES BY a)

c= a+b

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