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```        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```
7 years ago

SAGAR SINGH - IIT DELHI
879 Points
```										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.

All the best.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any                              query on askiitians forum and become an Elite      Expert        League            askiitian.

Sagar Singh
B.Tech, IIT Delhi

```
7 years ago
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