Question

If in a $\Delta ABC,\angle A={60}^{\circ },$ then find then value of $(1+\frac{a}{c}+\frac{b}{c})(1+\frac{c}{b}-\frac{a}{b})$

Answer 1

We need to find value of $(1+\frac{a}{c}+\frac{b}{c})(1+\frac{c}{b}-\frac{a}{b})$
$=\left(\frac{c+a+b}{c}\right)\left(\frac{b+c-a}{b}\right)$
$=\frac{{(b+c)}^2-a^2}{bc}$
$=\frac{(b^2+c^2-a^2+2bc)}{bc}$
$=\frac{b^2+c^2-a^2}{bc}+2=2\left(\frac{b^2+c^2-a^2}{2bc}\right)+2$
$=2\cos A+2$
$=2\cos {60}^o+2$
$=3$
$\therefore (1+\frac{a}{c}+\frac{b}{c})(1+\frac{c}{b}-\frac{a}{b})=3$