Question

Show that:
(i) $\frac{1-\sin 60°}{\cos 60°}=\frac{\tan 60°-1}{\tan 60°+1}$
(ii) $\frac{\cos 30°+\sin 60°}{1+\sin 30°+\cos 60°}=\cos 30°$

Answer 1

(i)
$$\begin{gathered} \mathrm{LHS}=\frac{1-\sin {60}^{\mathrm{o}}}{\cos {60}^{\mathrm{o}}}=\frac{1-{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}}=\frac{\left({\displaystyle \frac{2-\sqrt{3}}{2}}\right)}{{\displaystyle \frac{1}{2}}}=\left({\displaystyle \frac{2-\sqrt{3}}{2}}\right)\times 2=2-\sqrt{3} \\ \mathrm{RHS}=\frac{\tan {60}^{\mathrm{o}}-1}{\tan {60}^{\mathrm{o}}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{{\left(\sqrt{3}-1\right)}^2}{{\left(\sqrt{3}\right)}^2-1^2}=\frac{3+1-2\sqrt{3}}{3-1}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3} \end{gathered}$$

Hence, LHS = RHS

∴ $\frac{1-\sin {60}^{\mathrm{o}}}{\cos {60}^{\mathrm{o}}}=\frac{\tan {60}^{\mathrm{o}}-1}{\tan {60}^{\mathrm{o}}+1}$

(ii)
$$\begin{gathered} \mathrm{LHS}=\frac{\cos {30}^{\mathrm{o}}+\sin {60}^{\mathrm{o}}}{1+\sin {30}^{\mathrm{o}}+\cos {60}^{\mathrm{o}}}=\frac{\left({\displaystyle \frac{\sqrt{3}}{2}}+{\displaystyle \frac{\sqrt{3}}{2}}\right)}{\left(1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}\right)}=\frac{{\displaystyle \frac{\sqrt{3}+\sqrt{3}}{2}}}{{\displaystyle \frac{2+1+1}{2}}}=\frac{\sqrt{3}}{2} \\ \mathrm{Also},\mathrm{RHS}=\cos {30}^{\mathrm{o}}=\frac{\sqrt{3}}{2} \end{gathered}$$

Hence, LHS = RHS

∴ $\frac{\cos {30}^{\mathrm{o}}+\sin {60}^{\mathrm{o}}}{1+\sin {30}^{\mathrm{o}}+\cos {60}^{\mathrm{o}}}=\cos {30}^{\mathrm{o}}$1−sin60°cos60°