IIT JEE , AIEEE Forum - Mathematics , Trignometry

sanchay_1991

In

ABC the area of triangle is c² - (b - a)², then tan (C/2) is equal to:

(a) 4 (b) 1/4

(c) 1/2 (d) 1

rik_mad

abhijeet_0201

is the ans (b)1/4???????
my method is a bit long so if u ask then ill post it.

ankurgupta91

tan C/2 = (s-a)(s-b) / delta
= [(b+c-a)/2 * (a+c-b)/2 ] /delta [s=a+b+c/2]
= [c^2 -a^2-b^2+2ab] /4*delta
= c^2 -(b^2-a^2] /4*delta
nw delta =c^2 -(b^2-a^2]
tan C/2 =1/4
thanks.........

sanchay_1991

i agree with u ankur

thanksss...........

chirag

tan(C/2)={

/s(s-c)}

tan(C/2)=c^2-(b-a)^2/s(s-c)

tan(C/2)=(c+b-a)(c-b+a)/s(s-c)

tan(C/2)=2(s-a)2(s-b)/s(s-c)

tan(C/2)=4(s-a)(s-b)/s(s-c)

4{tan(C/2)}^2=tan(C/2)

tan(C/2)=1/4

Please rate me...........

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