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Asked by mailtome.aj86 | 26 Feb, 2024, 08:29: PM

a,b and c are sides of triangle

A, B and C are angles of triangle

a,b and c are sides opposite to angles A, B and C respectively.

By sine rule , we have

where k is constant .

Hence we get  ,

a = k sinA  ;   a2 = k2 sin2A .........................(1) ;

b = k sinB  ;   b2 = k2 sin2B .........................(2) ;

c = k sinC  ;   c2 = k2 sin2C .........................(3);

if a2 , b2 and c2 are in A.P., then we have ,

a2 + c2 = 2 b2 ..................(4)

if we subsititute a2 , b2 and c2 using eqn.(1), (2) and (3), then we rewrite eqn.(4) as

sin2A + sin2C = 2 sin2B ........................(5)

Hence from above equation,  sin2A , sin2B and sin2C are in A.P .

But sinA , sinB and sinC are not in A.P.

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If A, B and C are angles of triangle, then we have

tan(A+C) = tan(180-B) = - tanB

tanA + tanC = ( tanA tanC -1 ) tanB ......................(6)

if tanA , tanB and tanC are in A.P , then we get from eqn.(6) as

tanA tanC = 3

by sine rule , we have  sin A  = k a ...................(7)

by cosine rule , we have

..................(8)

From eqn.(7) and eqn.(8) , we get

..........................(9)

Similarly we get

..................(10)

It can be seen from eqn.(8), (9) and (10)  that even after applying the condition " 2b2 = c2 + a2 " ,

i.e., the condition if a2 , b2 and c2 are in A.P , we can not get " tanA tanC = 3 "

hence if a2 , b2 and c2 are in A.P , then tanA , tanB and tanC are not in A. P.

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