Question

Prove that: ${\displaystyle {\left(\frac{\cot 30°+1}{\cot 30°-1}\right)}^2=\frac{\sec 30°+1}{\sec 30°-1}}$

Answer 1

L.H.S. = ${\displaystyle {\left(\frac{\cot 30°+1}{\cot 30°-1}\right)}^2}$

= ${\displaystyle {\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)}^2}$

= ${\displaystyle {(\frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1})}^2}$

= ${\displaystyle \frac{{\left(\sqrt{3}\right)}^2+{\left(1\right)}^2+2\sqrt{3}}{{\left(\sqrt{3}\right)}^2+{\left(1\right)}^2-2\sqrt{3}}}$

= ${\displaystyle \frac{3+1+2\sqrt{3}}{3+1-2\sqrt{3}}}$

= ${\displaystyle \frac{4+2\sqrt{3}}{4-2\sqrt{3}}}$

= ${\displaystyle \frac{2(2+\sqrt{3})}{2(2-\sqrt{3})}}$

= ${\displaystyle \frac{2+\sqrt{3}}{2-\sqrt{3}}}$

= ${\displaystyle \frac{\frac{2}{\sqrt{3}}+1}{\frac{2}{\sqrt{3}}-1}}$

= ${\displaystyle \frac{\sec 30°+1}{\sec 30°-1}}$
= R.H.S.