1) if A+B+C=3.14, then prove that, cosA+cosB+cosC<3/2, where A,B,c are distinct.

2) if A+B+C=3.14,then prove that tan^2A/2+tan^2B/2+tan^2C/2>1

sir please solve these two question by using graph

Dear student,

Using CosA = [ (b² + c² – a²) / 2bc ]  etc. , in the relation  cosA  +  cosB  +  cosC  =  3/2  we obtain

[ (b² + c² – a²) / 2bc ] + [ (c² + a² – b²) / 2ac ] + [ (a² + b² – c²) / 2ab ] = 3/2

or   a.(b² + c² – a²)  +  b.(c² + a² – b²)  + c.(a² + b² – c²) = 3abc

or   a.(b² + c²)  +  b.(c² + a² )  + c.(a² + b² ) =  a³ + b³ + c³ + 3 abc

or   ab² + ac² +  bc² + ba²   + ca² + cb² =  a³ + b³ + c³ + 3 abc

Now subtract 6 abc from both the sides and rearraginf the terms we obtain

ab² + ac² - 2abc +  bc² + ba²  - 2abc  + ca² + cb²  - 2abc = a³ + b³ + c³ - 3 abc

or  a ( b - c )² + b ( c – a )² + c ( a – b )²  =  a³ + b³ + c³ - 3 abc

or  a ( b - c )² + b ( c – a )² + c ( a – b )²  =  (a + b + c) (a² + b² + c² – ab – bc – ca)                    ... ( 1 )

Multiply both the sides by 2 we get

Left hand side = 2a ( b - c )² + 2b ( c – a )² + 2c ( a – b )²

&   Right hand side = (a + b + c) (2a² + 2b² +2 c² – 2ab – 2bc – 2ca) 

or RHS = (a + b + c)  (a² + b² + c² – 2ab – 2bc – 2ca + a² + b² + c² ) 

 = ( a + b + c ) [ (a² + b² – 2ab) + ( b² + c² – 2 bc ) + (a² + c² - 2 ca) ]

 = ( a + b + c ) [ ( a – b )² + (b – c )² + ( c – a )² ]

Now eqn (1) becomes

2a ( b - c )² + 2b ( c – a )² + 2c ( a – b )² =  ( a + b + c ) [ ( a – b )² + (b – c )² + ( c – a )² ]

or (a - b - c) ( b - c )² + (b -a - c) ( c – a )² + ( c- a - b) ( a – b )² = 0       ... (2)

Now we know that sum of two sides of a triangle is greater than the third side. that is  (a - b - c) is not equal to 0 similarly (b - a - c) is not equal to 0 and also (c - a - b) is not equal to 0

Hence in eqn (2) squared terms must be equal to zero.

ie. ( b - c )² = 0  or  b = c . Similarly c = a

Hence a = b = c

 i.e. The triangle is Equilateral.

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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