(1) In a triangle ABC,angle A=pi/2, then tanC/2 is equal to

ans = (a)a-c/2b

(b)a-b/2c

(c)a-c/b

(d)a-b/c

let the triangle be ABC right angled at A as said

here sicC=c/a

cosC=b/a

hence tanC=c/b

 

   -1/tanC

=root of c²+b²   -b/c

which is a-c/b

rate if useful