Question

Show that:

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

(ii) $\frac{\cos {30}^{\circ }+\sin {60}^{\circ }}{1+\sin {30}^{\circ }+\cos {60}^{\circ }}=\cos {30}^{\circ }$

Answer 1

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

L.H.S $=\frac{1-\sin {60}^{\circ }}{\cos {60}^{\circ }}$

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

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

$=\frac{2-\sqrt{3}}{2}\times \frac{2}{1}$

$=2-\sqrt{3}$

R.H.S $=\frac{\tan {60}^{\circ }-1}{\tan {60}^{\circ }+1}$

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

$=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$$\left[\text{on~rationalising}\right]$

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

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

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

$=2-\sqrt{3}$

L.H.S $=$ R.H.S

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

(ii) $\frac{\cos {30}^{\circ }+\sin {60}^{\circ }}{1+\sin {30}^{\circ }+\cos {60}^{\circ }}=\cos {30}^{\circ }$

L.H.S $=\frac{\cos {30}^{\circ }+\sin {60}^{\circ }}{1+\sin {30}^{\circ }+\cos {60}^{\circ }}$

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

$=\frac{(\frac{2\sqrt{3}}{2})}{2}$

$=\frac{\sqrt{3}}{2}$

R.H.S $=\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$

L.H.S $=$ R.H.S

$\therefore \frac{\cos {30}^{\circ }+\sin {60}^{\circ }}{1+\sin {30}^{\circ }+\cos {60}^{\circ }}=\cos {30}^{\circ }$