Question

in the given triangle ABC, prove that​

Answer 1

Third answer refers to the attachment

รσℓµƭเσɳ →

1st To prove ->

a( b cosC - c cosB) = b² - c²

Proof →

We know that ,In cosine formula :-

$\cos(c) = \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab}$

And in

$\cos(b) = \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2ac}$

Now substituting these values in LHS :-

LHS →

a ( b cosC - c cosB)

$a(b. \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} - c. \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2ac} )$

$( \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2} - \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2}$

Taking 2 as LCM

$( \frac{ {a}^{2} + {b}^{2} - {c}^{2} - {a}^{2} - {c}^{2} + {b}^{2} }{2}$

$( \frac{2 {b}^{2} - 2 {c}^{2} }{2} )$

${b}^{2} - {c}^{2} = rhs$

$hence \: proved$

2nd To prove →

a²( cos²B - cos²C) + b²( cos²C - cos²A) + c²( cos²A - cos²B) = 0

Proof →

Taking LHS :-

a²( cos²B - cos²C ) + b²( cos²C - cos²A ) + c²( cos²A - cos²B)

Now we know that cos²a = ( 1- sin²a) so similarly changing all cos angles into sine .

→a²[ ( 1-sin²B) - ( 1-sin²C)] +b²[( 1-sin²C) - (1- sin²A)]+ c²[( 1-sin²A)-(1- sin²B)]

→a² ( sin²C - sin²B ) + b²( sin²A - sin²C) +c²( sin²B-sin²A)

As we know that a =k SinA , b= ksinB , c= ksinC ,So :-

→ k(Sin²Asin²C - sin²Asin²B + sin²Bsin²A- sin²Bsin²C +sin²Csin²B - sin²Csin²A )

→k(0)

→0 = RHS

$hence \: proved$