Question

 Show that:

(i)${\displaystyle \frac{1-{\sin 60}^0}{{\cos 60}^0}=\frac{{\tan 60}^0-1}{{\tan 60}^0+1}}$

Answer 1

LHS=${\displaystyle \frac{1-{\sin 60}^0}{{\cos 60}^0}=\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\left(\frac{2-\sqrt{3}}{2}\right)}{\frac{1}{2}}=\left(\frac{2-\sqrt{3}}{2}\right)\times 2=2-\sqrt{3}}$

RHS=${\displaystyle \frac{{\tan 60}^0-1}{{\tan 60}^0+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{{(\sqrt{3}-1)}^2}{{\left(\sqrt{3}\right)}^2-1^2}=\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}}$

Hence, LHS = RHS

${\displaystyle \therefore \frac{1-{\sin 60}^0}{{\cos 60}^0}=\frac{{\tan 60}^0-1}{{\tan 60}^0+1}}$