In the triangle ABC, if 5 (b square + C square - a square) = 6BC, find sin2a + 2sina square / 1 + Tana Please~~ | Math enthusiast | Mathematical art

In the triangle ABC, if 5 (b square + C square - a square) = 6BC, find sin2a + 2sina square / 1 + Tana Please~~

5(b²+c²-a²)=6bc
According to the cosine theorem a & # 178; = B & # 178; + C & # 178; - 2bccosa, a is the angle of A
The above formula can be reduced to 10bccosa = 6BC to get cosa = 3 / 5 and Sina = 4 / 5
such
(sin2A+2sin²A)/(1+tanA)=2sinA(cosA+sinA)/[(cosA+sinA)/(cosA)]
=2sinAcosA=24/25

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