Question

Slove these 2 Questions :

1) If b cos(∅ + 120°) = c cos(∅ + 240°) , then prove that , b-c = -(b+c)√3 tan∅ .
2) Prove that ,
$[tex]sec \: \alpha + tan \: \alpha = tan( \frac{\pi}{4} + \frac{ \alpha }{2} )$[/tex]
Don't give wrong answer ! ​

Answer 1

Solution 1 :-

$\sf \: b \: cos( \theta + 120 \degree) = c \: cos( \theta + 240 \degree)$

$\implies \sf \: \dfrac{b}{c} = \dfrac{cos( \theta + 240 \degree)}{cos( \theta + 120 \degree)}$

$\implies \sf \: \dfrac{b + c}{b - c} = \dfrac{cos( \theta + 240 \degree) + cos( \theta + 120 \degree)}{cos( \theta + 240 \degree) - cos( \theta + 120 \degree)} \bigg \{componendo \: and \: dividendo \bigg \}$

$\implies \sf \: \dfrac{b + c}{b - c} = \dfrac{2 \: cos \dfrac{( \theta + 240 \degree + \theta + 120 \degree)}{2} cos \dfrac{( \theta + 240 \degree - \theta - 120 \degree)}{2} }{2 \: sin \dfrac{( \theta + 240 \degree + \theta + 120 \degree)}{2} sin \: \dfrac{( \theta + 120 \degree - \theta - 240 \degree)}{2} }$

$\implies \sf \: \dfrac{b + c}{b - c} = \dfrac{cos(180 \degree + \theta)cos \: 60 \degree}{sin(180 \degree + \theta)sin( - 60 \degree)}$

$\implies \sf \: \dfrac{b + c}{b - c} = - cot(180 \degree + \theta)cot60 \degree$

$\implies \sf \: \dfrac{b + c}{b - c} = - cot \theta \times \dfrac{1}{ \sqrt{3} }$

$\implies \sf \: \dfrac{b + c}{b - c} = - \dfrac{1}{ \sqrt{3} \: tan \theta}$

$\implies \sf \: b - c = - (b + c) \sqrt{3} \: tan \theta \: \big \{proved \big \}$

Solution 2 :-

Refer to the attachment .