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Question

If cosA+cosB=4·sin2C2 then prove that the sides a,c,b of a triangle ABC is in A.P


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Solution

Prove the given expression:

Given,

cosA+cosB=4·sin2C22cosA+B2cosA-B2=4·sin2C2

Sum of angles of triangle =A+B+C=180°

So, A+B=180°-C

2cos180°-C2cosA-B2=4·sin2C22cos90°-C2cosA-B2=4·sin2C22sinC2cosA-B2=4·sin2C2cosA-B2=2·sinC2

Multiply both sides by cosC2 we get

cosC2cosA-B2=2·sinC2cosC2cosC2cosA-B2=sinCsin2θ=2sinθcosθcos180°-A-B2cosA-B2=sinCsinA+B2cosA-B2=sinC

Multiplying both side by 2

2sinA+B2cosA-B2=2sinCsinA+sinB=2sinC....(1)

We know that

sinAa=sinBb=sinCc=t[let'ssay]sinA=at,sinB=btandsinC=ctat+bt=2ct(using(1))a+b=2c

So, the sides a,c,b of a triangle ABC are in A.P.

Hence proved.


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