Question

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

Answer 1

Hint: For this question, use the following values: $\sin 30{}^\circ =\dfrac{1}{2}$, $\tan 45{}^\circ =1$, \[\csc 60{}^\circ =\dfrac{2}{\sqrt{3}}\], \[\sec 30{}^\circ =\dfrac{2}{\sqrt{3}}\], $\cos 60{}^\circ =\dfrac{1}{2}$, and \[\cot 45{}^\circ =1\]. Substitute these values in the given expression: $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\csc 60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ +\cot 45{}^\circ }$. Simplify this and arrive at the final answer.

Complete step-by-step answer:

In this question, we need to find the value of the following expression: $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\csc 60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ +\cot 45{}^\circ }$
In this question, we will use trigonometric functions.

Let us first define what trigonometric functions actually are.

In mathematics, the trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. They are widely used in all sciences that are related to geometry, such as navigation, solid mechanics, celestial mechanics, geodesy, and many others.

Here, the angles given are standard angles and we directly know the values of the trigonometric functions for these angles.

Now, let us find the values of the trigonometric functions for these angles.

We already know that $\sin 30{}^\circ =\dfrac{1}{2}$

We also know that $\tan 45{}^\circ =1$

Now, we know that for any angle, csc or cosecant is the reciprocal of sine of the angle.

We also know that \[\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}\].

So, using the above property and the expression \[\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}\], we will get the following:

\[\csc 60{}^\circ =\dfrac{2}{\sqrt{3}}\]

Similarly, we know that for any angle, sec or secant is the reciprocal of cosine of the angle.

We also know that \[\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}\].

So, using the above property and the expression \[\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}\], we will get the following:

\[\sec 30{}^\circ =\dfrac{2}{\sqrt{3}}\]

We know that $\cos 60{}^\circ =\dfrac{1}{2}$.

Now, we know that for any angle, cot is the reciprocal of tan of the angle.

We also know that \[\tan 45{}^\circ =1\].

So, using the above property and the expression \[\tan 45{}^\circ =1\], we will get the following:

\[\cot 45{}^\circ =1\]

Putting all these values in $\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\csc 60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ +\cot 45{}^\circ }$, we will get the following:

$\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\csc 60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ +\cot 45{}^\circ }=\dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}}+\dfrac{1}{2}+1}$

$\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\csc 60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ +\cot 45{}^\circ }=\dfrac{\dfrac{3}{2}-\dfrac{2}{\sqrt{3}}}{\dfrac{3}{2}+\dfrac{2}{\sqrt{3}}}$

Multiplying numerator and denominator by $2\sqrt{3}$ to get the following:

$\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\csc 60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ +\cot 45{}^\circ }=\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}$

Rationalize denominator by multiplying the numerator and denominator by $3\sqrt{3}-4$ to get the following:

$\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\csc 60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ +\cot 45{}^\circ }=\dfrac{43-24\sqrt{3}}{11}$

$\dfrac{\sin 30{}^\circ +\tan 45{}^\circ -\csc 60{}^\circ }{\sec 30{}^\circ +\cos 60{}^\circ +\cot 45{}^\circ }=\dfrac{43-24\sqrt{3}}{11}$

This is the final answer.

Note: In this question, we need to know about the values of trigonometric functions of various standard angles. Also you need to know about rationalizing. To rationalize, we will multiply numerator and denominator by such a number that helps to remove the root from the denominator and move to the numerator.