Question

Prove that in ∆ :


a(bCosC - C CosB) = b² - C²


Prove that in ∆ :


a(bCosC - C CosB) = b² - C²

Prove that in ∆ :


a(bCosC - C CosB) = b² - C²

Answer 1

Question :

Prove that in ∆ :

a(bCosC - c CosB) = b² - C²

Step by Step Explanation :

Equation : a(bCosC - c CosB) = b² - C²

This Equation can Written as :

✏ (bCosC - c CosB) = b² - c² ÷ a

Now, Solving this Question!

$\bf{\implies{\red{ (b \: CosC - c \: CosB) = \frac{ {b}^{2} - {c}^{2} }{a} }}} \\ \\ \bf{\implies{\red{{b[ \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} - \: c[ \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2ac}}}}} \\ \\ \bf{\red{ \cancel{b}[ \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2a \cancel{b}} - \: \cancel{c}[ \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2a \cancel{c}}}} \\ \\ \bf{\red{[\frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2a} - \frac{ {a}^{2} + {c}^{2} - {b}^{2} }{2a}}}) \\ \\ \bf{\red{ \frac{ {a}^{2} + {b}^{2} - {c}^{2} - {a}^{2} - {c}^{2} + {b}^{2} }{2a}}} \\ \\ \bf{\red{ \frac{ \cancel{{a}^{2}} + {b}^{2} - {c}^{2} - \cancel{ {a}^{2}} - {c}^{2} + {b}^{2} }{2a}}} \\ \\ \bf{\red{( \frac{2 {b}^{2} - 2 {c}^{2} }{2a}})} \\ \\ \bf{\red{ \frac{2( {b}^{2} - {c}^{2} ) }{2a}}} \\ \\ \bf{\red{ \frac{ \cancel{2}( {b}^{2} - {c}^{2} ) }{ \cancel{2}a}}} \\ \\ \bf{\green{ \frac{( {b}^{2} - {c}^{2}) }{a}}}$

Hence, it's Proved!