Question

Prove in any triangle, r_1=stan(A/2)

Answer 1

Answer:

On using the sine rule we know, a/sin A = b/sin B = c/sin C = k a/sin A = k Therefore, a = k sin A Now, similarly b = k sin B And c = k sin C ...(i) As we know, a + b = k(sin A + sin B) ...(ii) Let us consider the LHS c/(a + b)  On substituting the equation (i) and (ii) in above equation, k sin C/k(sin A + sin B) = sin C/(sin A + sin B) ...(iii) On applying half angle rule, sin C = 2 sin(C/2) cos(C/2) ...(iv) And as we know sin A + sin B = 2 sin(A + B)/2 cos(A - B)/2 ...(v) On substituting the above equation (iv) and (v) in equation (iii) As we know, cos (A + B)/2 = cos (A/2 + B/2) = cos A/2 cos B/2 + sin A/2 sin B/2 cos (A – B)/2 = cos (A/2 – B/2) = cos A/2 cos B/2 – sin A/2 sin B/2 On substituting the above equations in equation (vi), Now, let us divide the numerator and denominator by cos A/2 cos B/2, = RHS Thus proved.Read more on Sarthaks.com - https://www.sarthaks.com/657146/in-any-triangle-abc-prove-the-c-a-b-1-tan-a-2-tan-b-2-1-tan-a-2-tan-b-2?show=657219#a657219