Question

Find the value of $\sin {60^0} + \cos {30^0} - \tan {30^0}$

Answer 1

Hint: First, the given values are in the form of the trigonometric values because it contains the trigonometric identities of sine, cosine, and tangent.
The trigonometric functions are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.

Complete step-by-step solution:
Since from the given, we have the two values of the sine, cosine, and tangent as $\sin {60^0}$ , $\cos {30^0}$ and $\tan {30^0}$. We need to find the exact values of the given functions.
Let us start with the sine table in the trigonometric functions with respect to the corresponding angles

Angle in degrees	\[0^\circ \]	\[30^\circ \]	\[45^\circ \]	\[60^\circ \]	\[90^\circ \]
\[\sin \]	\[0\]	\[\dfrac{1}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{{\sqrt 3 }}{2}\]	\[1\]

Similarly cosine table in the trigonometric functions with respect to the corresponding angles

Angle in degrees	\[0^\circ \]	\[30^\circ \]	\[45^\circ \]	\[60^\circ \]	\[90^\circ \]
\[\cos \]	\[1\]	\[\dfrac{{\sqrt 3 }}{2}\]	\[\dfrac{1}{{\sqrt 2 }}\]	\[\dfrac{1}{2}\]	\[0\]

And also we know that the relation of the sine, cosine, and tangent is $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $ and hence we get the tangent table as

Angle in degrees	\[0^\circ \]	\[30^\circ \]	\[45^\circ \]	\[60^\circ \]	\[90^\circ \]
$\tan $	\[0\]	\[\dfrac{1}{{\sqrt 3 }}\]	\[1\]	\[\sqrt 3 \]	undefined

Hence the value of $\sin {60^0} + \cos {30^0} - \tan {30^0}$ can be easily found in the given tables.
Since $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$ in the first table, $\cos {30^0} = \dfrac{{\sqrt 3 }}{2}$ in the second table, and thus $\tan {30^0} = \dfrac{1}{{\sqrt 3 }}$ in the final table.
Thus using the addition and subtraction operation we get $\sin {60^0} + \cos {30^0} - \tan {30^0} = \dfrac{{\sqrt 3 }}{2} + \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 3 }}$
Since we know that $\dfrac{1}{2} + \dfrac{1}{2} = 1$ and then we get $\sin {60^0} + \cos {30^0} - \tan {30^0} = \sqrt 3 - \dfrac{1}{{\sqrt 3 }}$
Now making use of the cross multiplication method we get $\sin {60^0} + \cos {30^0} - \tan {30^0} = \dfrac{{(\sqrt 3 \times \sqrt 3 ) - 1}}{{\sqrt 3 }}$
Since $\sqrt 3 \times \sqrt 3 = 3$ and then we get $\sin {60^0} + \cos {30^0} - \tan {30^0} = \dfrac{{3 - 1}}{{\sqrt 3 }}$
Hence by subtraction, we get $\sin {60^0} + \cos {30^0} - \tan {30^0} = \dfrac{2}{{\sqrt 3 }}$ which is the required answer.

Note: Both values of the $\sin {150^0} = \sin {30^0}$ are the same because the reference angle for the $150$ is equal to the $30$ triangle formed in the unit circle.
The given angle is referenced in the form when the perpendicular is dropped from the unit circle to the x-axis, which forms a right triangle.
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan $and $\tan = \dfrac{1}{{\cot }}$
Also, it will be represented in the angles corresponding to the given trigonometric values. The highest angle is \[{360^0}\] a degree or it can be also expressed as ${0^0}$ in the circle.