Question
Prove that: $\left(\frac{\cot 30^{\circ}+1}{\cot 30^{\circ}-1}\right)^2=\frac{\sec 30^{\circ}+1}{\sec 30^{\circ}-1}$
✓
Answer
$ \text { L.H.S. }=\left(\frac{\cot 30^{\circ}+1}{\cot 30^{\circ}-1}\right)^2$
$ =\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2$
$ =\left(\frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\right)^2$
$ =\frac{(\sqrt{3})^2+(1)^2+2 \sqrt{3}}{(\sqrt{3})^2+(1)^2-2 \sqrt{3}}$
$ =\frac{3+1+2 \sqrt{3}}{3+1-2 \sqrt{3}}$
$ =\frac{4+2 \sqrt{3}}{4-2 \sqrt{3}}$
$ =\frac{2(2+\sqrt{3})}{2(2-\sqrt{3})}$
$ =\frac{2+\sqrt{3}}{2-\sqrt{3}}$
$ =\frac{\frac{2}{\sqrt{3}}+1}{\frac{2}{\sqrt{3}}-1}$
$ =\frac{\text { sec } 30^{\circ}+1}{\text { Sec } 30^{\circ}-1}$
$ =\text { R.H.S. }$
$ =\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)^2$
$ =\left(\frac{\sqrt{3}+1}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\right)^2$
$ =\frac{(\sqrt{3})^2+(1)^2+2 \sqrt{3}}{(\sqrt{3})^2+(1)^2-2 \sqrt{3}}$
$ =\frac{3+1+2 \sqrt{3}}{3+1-2 \sqrt{3}}$
$ =\frac{4+2 \sqrt{3}}{4-2 \sqrt{3}}$
$ =\frac{2(2+\sqrt{3})}{2(2-\sqrt{3})}$
$ =\frac{2+\sqrt{3}}{2-\sqrt{3}}$
$ =\frac{\frac{2}{\sqrt{3}}+1}{\frac{2}{\sqrt{3}}-1}$
$ =\frac{\text { sec } 30^{\circ}+1}{\text { Sec } 30^{\circ}-1}$
$ =\text { R.H.S. }$
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