Question

Find the value of the following:
$\frac{\sin {30}^o+\tan {45}^o-\csc {60}^o}{\sec {30}^o+\cos {60}^o+\cot {45}^o}$

Answer 1

We know that the values of the trignometric functions:

$\sin {30}^0=\cos {60}^0=\frac{1}{2}$,

$\tan {45}^0=\cot {45}^0=1$ and,

$\csc {60}^0=\sec {30}^0=\frac{2}{\sqrt{3}}$

Therefore, the value of $\frac{\sin {30}^0+\tan {45}^0-\csc {60}^0}{\sec {30}^0+\cos {60}^0+\cot {45}^0}$ is:

$\frac{\sin {30}^0+\tan {45}^0-\csc {60}^0}{\sec {30}^0+\cos {60}^0+\cot {45}^0}=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\frac{4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}=\frac{\frac{3\sqrt{3}-4}{2\sqrt{3}}}{\frac{4+3\sqrt{3}}{2\sqrt{3}}}=\frac{3\sqrt{3}-4}{2\sqrt{3}}\times \frac{2\sqrt{3}}{4+3\sqrt{3}}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}$

Hence, $\frac{\sin {30}^0+\tan {45}^0-\csc {60}^0}{\sec {30}^0+\cos {60}^0+\cot {45}^0}=\frac{3\sqrt{3}-4}{3\sqrt{3}+4}$.