JEE Class main Answered

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.