In $\Delta ABC$, if $A=60$ then the value of $(1+\frac ac +\frac bc)(1+\frac cb -\frac ab)$

Question

So I cheated a bit in this question. Since there are no other conditions given, I assumed the triangle to be equilateral, and so the answer ends up being 3. Now I don’t have the right answer with me, but the options were

3,2,1,4

Is my answer right, and what is the generalized way to solve this?

Answer 1

Note,

$$\left(1+\frac ac +\frac bc\right)\left(1+\frac cb -\frac ab\right)$$ $$=\frac 1{bc}(c+a+b)(b+c-a)=\frac {(c+b)^2-a^2}{bc}$$ $$=\frac {c^2+b^2-a^2 +2bc}{bc}=2\cos A +2 = 2\cdot\frac12 + 2 = 3$$

Answer 2

We reach at $$\dfrac{(b+c)^2-a^2}{bc}$$

Now use $$2bc\cos60^\circ=b^2+c^2-a^2$$ to eliminate $$b^2+c^2-a^2$$

Answer 3

You can use the law of sines; $\frac{a}{c} = \frac{\sin A}{\sin C}$, together with the fact that $A+B+C=\pi$, and simplify.