In ΔABC the sides opposite to angles A,B,C are denoted by a,b,c respectively.
If cosA+cosB+2cosC=2 then the sides of the ΔABC are in?
A
AP
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B
GP
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C
HP
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D
none of these
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Solution
The correct option is A AP Given, cosA+2cosB+cosC=2⇒cosA+cosC=2(1−cosB) ⇒2cos(A+C2)cos(A−C2)=2×2sin2B2 ⇒2sinB2cos(A−C2)=2×2sinB2cos(A+C2) ⇒cos(A−C2)=2cos(A+C2) ⇒cosA2cosC2=3sinA2sinC2 ⇒tanA2tanC2=3 ⇒√(s−b)(s−c)s(s−a)×√(s−a)(s−b)s(s−c)=13 ⇒3(s−b)=s⇒2s=3b ⇒a+b+c=3b⇒b=a+c2 Hence a,b,c are in A.P