Question
If $A = 30^\circ;$show that:$\frac{1-\cos 2 A}{\sin 2 A}=\tan A$
✓
Answer
Given that $A = 30^\circ$
$\text { LHS }=\frac{1-\cos 2 A }{\sin 2 A }$
$=\frac{1-\cos 2\left(30^{\circ}\right)}{\sin 2\left(30^{\circ}\right)}$
$=\frac{1-\frac{1}{2}}{\frac{\sqrt{3}}{2}}$
$=\frac{1}{\sqrt{3}}$
$\text { RHS }=\tan A$
$=\tan 30^{\circ}$
$=\frac{1}{\sqrt{3}}$
$\text { LHS }=\text { RHS }$
$\text { LHS }=\frac{1-\cos 2 A }{\sin 2 A }$
$=\frac{1-\cos 2\left(30^{\circ}\right)}{\sin 2\left(30^{\circ}\right)}$
$=\frac{1-\frac{1}{2}}{\frac{\sqrt{3}}{2}}$
$=\frac{1}{\sqrt{3}}$
$\text { RHS }=\tan A$
$=\tan 30^{\circ}$
$=\frac{1}{\sqrt{3}}$
$\text { LHS }=\text { RHS }$
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