How does $\lambda \sqrt{\frac{1+v/c}{1-v/c}}$ become $\lambda \frac{1+v/c}{\sqrt{1-v^2/c^2}}$?

Question

I was reading a physics text and came across this equation :

$$\large \lambda \sqrt{\frac{1+ \frac vc}{1-\frac vc}} = \lambda \frac{1+\frac vc}{\sqrt{1-\frac{v^2}{c^2}}}$$

I am confused as to how they said both were equal. I'm not sure how to start this to make these equal. I would assume that I need to square some things in the square root, but I am unsure.

Answer 1

In such questions with a + in numerator and - in denominator or vice versa, we can multiply by $\sqrt{1+\frac{v}{c}}$ in the numerator and denominator to get rid of one of the square roots that usually makes problems easier

Now it is easy to see that after multiplying by $\sqrt{1+\frac{v}{c}}$ in numerator, we get numerator as $\left(\sqrt{1+\frac{v}{c}}\right)^2 = 1+\frac{v}{c}$ and in the denominator after multiplying by $\sqrt{1+\frac{v}{c}}\;$ we get denominator as $$\sqrt{1-\frac{v}{c}} \cdot \sqrt{1+\frac{v}{c}} = \sqrt{1-\frac{v^2}{c^2}}$$ (remembering that $(a-b)(a+b) = a^2 - b^2$)