Question
Evaluate: $\frac{\sin 30^{\circ}+\tan 45^{\circ}-\ cosec 60^{\circ} }{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
✓
Answer
We have $\frac{\sin 30^{\circ}+\tan 45^{\circ}- \ cosec 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$
after putting values,we get
$=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}$
$=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}$
$=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4}$ Rationalise it, we get
$=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4} \times \frac{3 \sqrt{3}-4}{3 \sqrt{3}-4}$
$=\frac{(3 \sqrt{3}-4)^{2}}{(3 \sqrt{3})^{2}-(4)^{2}}$
$=\frac{27+16-24 \sqrt{3}}{27-16}$
$=\frac{43-24 \sqrt{3}}{11}$
after putting values,we get
$=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}$
$=\frac{\frac{3}{2}-\frac{2}{\sqrt{3}}}{\frac{3}{2}+\frac{2}{\sqrt{3}}}$
$=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4}$ Rationalise it, we get
$=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4} \times \frac{3 \sqrt{3}-4}{3 \sqrt{3}-4}$
$=\frac{(3 \sqrt{3}-4)^{2}}{(3 \sqrt{3})^{2}-(4)^{2}}$
$=\frac{27+16-24 \sqrt{3}}{27-16}$
$=\frac{43-24 \sqrt{3}}{11}$
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