Question

Evaluate the following expression $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$.

Answer 1

Hint: Using trigonometric standard angles table, write the expression in the values and then further rationalize it to get the desired answer.

Complete step-by-step answer:
In this question, we have been given an expression, $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$ and we have to evaluate or find its value. If we look at the given expression, then we can see that the angels have their trigonometric ratios as the standard angles. Generally, the values of trigonometric standard angles are usually written in a trigonometric standard angles table. The table is shown below.

| $0{}^\circ$ | $30{}^\circ$ | $45{}^\circ$ | $60{}^\circ$ | $90{}^\circ$
---|---|---|---|---|---
Sin| 0| $\dfrac{1}{2}$ | $\dfrac{1}{\sqrt{2}}$ | $\dfrac{\sqrt{3}}{2}$ | 1
Cos| 1| $\dfrac{\sqrt{3}}{2}$| $\dfrac{1}{\sqrt{2}}$| $\dfrac{1}{2}$| 0
Tan| 0| $\dfrac{1}{\sqrt{3}}$ | 1| $\sqrt{3}$ | $\infty$
Cosec| $\infty$| 2| $\sqrt{2}$ | $\dfrac{2}{\sqrt{3}}$ | 1
Sec| 1| $\dfrac{2}{\sqrt{3}}$| $\sqrt{2}$| 2| $\infty$
cot| $\infty$| $\sqrt{3}$| 1| $\dfrac{1}{\sqrt{3}}$| 0

To solve the given expression, we need the values of only some of the trigonometric standard angles that are mentioned in the question given to us. So, according to the question, we only need the values of $\cos 45{}^\circ ,\sec 30{}^\circ$ and $\operatorname{cosec}30{}^\circ$. We can see from the trigonometric standard angles table given above that the value of $\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\sec 30{}^\circ =\dfrac{2}{\sqrt{3}}$ and $\operatorname{cosec}30{}^\circ =2$.
So, on substituting these values in the given expression, we get, $\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}$.
Now, we will multiply both the numerator and the denominator by $2\sqrt{3}$. So, we get, $\dfrac{\dfrac{1}{\sqrt{2}}\times 2\sqrt{3}}{\dfrac{2}{\sqrt{3}}\times 2\sqrt{3}+2\times 2\sqrt{3}}$
On simplifying the above expression, we get,
$\dfrac{\sqrt{6}}{4+4\sqrt{3}}=\dfrac{\sqrt{6}}{4\left( \sqrt{3}+1 \right)}$
Now we will rationalize the function by multiplying with $\left( \sqrt{3}-1 \right)$ to both the numerator and the denominator. So, we get,
$\dfrac{\sqrt{6}}{4\left( \sqrt{3}+1 \right)}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}$
Now, on multiplication and further simplification, we get,
$\dfrac{\sqrt{6}\left( \sqrt{3}-1 \right)}{4\times \left( \sqrt{3}-1 \right)}=\dfrac{\sqrt{6}\left( \sqrt{3}-1 \right)}{8}$
Hence, the value of the given expression, $\dfrac{\cos 45{}^\circ }{\sec 30{}^\circ +\operatorname{cosec}30{}^\circ }$ is $\dfrac{\sqrt{6}\left( \sqrt{3}-1 \right)}{8}$.

Note: While doing rationalization, if the fraction is in the form of $\dfrac{c}{a+\sqrt{6}}$ then we can rationalize it by multiplying the numerator and the denominator by $a-\sqrt{b}$ and simplify it further.