In a triangle ABC if tanA2 = p and tan B2 = q, prove that sinC=2(p+q)(1−pq)(1+p2)(1+q2)
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Solution
We have R.H.S =2(p+q)(1−pq)(1+p2)(1+q2) 2[tan(A/2)+tan(B/2)][1−tan(A/2)tan(B/2)][1+tan2(A/2)][1+tan(B/2)] 2sin12(A+B)cos12(A+B)(sin212A+cos212A)(sin212B+cos212B) =2sin12(A+B)cos12(A+B) =sin(A+B)=sin(π−C)=sinC