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UPSC NDA II 2015 Mathematics Solved Paper

For the next two (2) items that follow :

Consider a triangle ABC satisfying

2a\sin^2\left(\frac{C}{2}\right)+2c\sin^2\left(\frac{A}{2}\right)=2a+2c-3b

[NDA II 2015]

Question : 38 of 120

Marks: +1, -0

sin A,sinB,sin C are in

Solution:

As we have already proven

2b=a+c

as-

a = R sinA

b = R sinB

C = R SinC

\Rightarrow 2(R\sin B)=R\sin A+R\sin C

\Rightarrow 2R(\sin B)=R(\sin A+\sin C)

2\sin B=\sin A+\sin C

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