Question

Answer this 23th question rightly and I'll mark it as brainliest.​

Answer 1

Required Answer:-

Given to prove:

• $\sf \small \dfrac{c - b\cos(A) }{b - c \cos(A) } = \dfrac{ \cos(B) }{ \cos(C) }$

Proof:

By cosine formula, we get,

$\sf \mapsto \cos(A) = \small \dfrac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc}$

$\sf \mapsto \cos(B) = \small \dfrac{ {a}^{2} + {c}^{2} - {b}^{2} }{2ac}$

$\sf \mapsto \cos(C) = \small \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab}$

Taking LHS,

$\sf \small \dfrac{c - b\cos(A) }{b - c \cos(A) }$

$\sf = \dfrac{c - b \bigg( \dfrac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} \bigg)}{b - c \bigg( \dfrac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} \bigg)}$

$\sf = \dfrac{c - \bigg( \dfrac{ {b}^{2} + {c}^{2} - {a}^{2} }{2c} \bigg)}{b - \bigg( \dfrac{ {b}^{2} + {c}^{2} - {a}^{2} }{2b} \bigg)}$

$\sf = \dfrac{\bigg( \dfrac{ 2 {c}^{2} - ({b}^{2} + {c}^{2} - {a}^{2}) }{2c} \bigg)}{\bigg( \dfrac{2 {b}^{2} - ( {b}^{2} + {c}^{2} - {a}^{2} )}{2b} \bigg)}$

$\sf = \dfrac{\bigg( \dfrac{ {a}^{2} + {c}^{2} - {b}^{2} }{2c} \bigg)}{\bigg(\dfrac{ {a}^{2} + {b}^{2} - {c}^{2}}{2b} \bigg)}$

$\sf = \dfrac{\bigg( \dfrac{ {a}^{2} + {c}^{2} - {b}^{2} }{2ac} \bigg)}{\bigg(\dfrac{ {a}^{2} + {b}^{2} - {c}^{2}}{2ab} \bigg)}$

$\sf = \dfrac{ \cos(B) }{ \cos(C) }$

Taking RHS,

$\sf = \dfrac{ \cos(B) }{ \cos(C) }$

Hence, LHS = RHS (Proved)