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Question:

Let PQR be a triangle of area Δ with a=2, b=72 and c=52, where a, b and c are the lengths of the sides of the triangle opposite to the angles at P, Q and R respectively. Then 2sinP−sin2P2sinP+sin2P equals

454Δ

(454Δ)²

34Δ

(34Δ)²

Solution:

2sinP−sin2P2sinP+sin2P=2sinP−2sinPcosP2sinP+2sinPcosP=1−cosP1+cosP=2sin²P22cos²P2=tan²P2=(s−b)(s−c)s(s−a)=((s−b)(s−c))²Δ²[∵Δ=√s(s−a)(s−b)(s−c)]=((12)(32))²Δ²[s=a+b+c2]=(34Δ)²