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Grade 11,

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1) if A+B+C=3.14, then prove that, cosA+cosB+cosC 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

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

Grade:11

1 Answers

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

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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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

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