Question

The value of $\sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ +\cot 90{}^\circ$ is equal to
A. $\dfrac{5\sqrt{3}-6}{6}$
B. $\dfrac{-6+7\sqrt{3}}{6}$
C. $0$
D. $2$

Answer 1

In this problem we need to calculate the value of the given expression. We can observe that the given expression has a trigonometric value for different angles and the value of trigonometric ratios of different angles is also known. So, we will list all the values of trigonometric ratios which are in the given expression and add or subtract them one by one according to the given expression to get the required value.

Complete step by step solution:
Given expression is $\sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ +\cot 90{}^\circ$.
In the above expression we have the values of $\sin 0{}^\circ$, $\cos 30{}^\circ$, $\tan 45{}^\circ$, $\csc 60{}^\circ$, $\cot 90{}^\circ$.
We can observe that the all the above values are simply obtained by the trigonometric table. From the trigonometric table the values of the above ratios are given by
$\sin 0{}^\circ =0$
$\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$
$\tan 45{}^\circ =1$
$\csc 60{}^\circ =\dfrac{2}{\sqrt{3}}$
$\cot 90{}^\circ =0$
Now the value of $\sin 0{}^\circ +\cos 30{}^\circ$ will be
$\begin{aligned} & \sin 0{}^\circ +\cos 30{}^\circ =0+\dfrac{\sqrt{3}}{2} \\\ & \Rightarrow \sin 0{}^\circ +\cos 30{}^\circ =\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$
Subtract the value of $\tan 45{}^\circ$ from both sides of the above equation, then we will have
$\sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ =\dfrac{\sqrt{3}}{2}-1$
Simplifying the above equation by taking LCM on left hand side, then we will get
$\sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ =\dfrac{\sqrt{3}-2}{2}$
Now add $\csc 60{}^\circ$ on both sides of the above equation, then we will have
$\sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ =\dfrac{\sqrt{3}-2}{2}+\dfrac{2}{\sqrt{3}}$
Simplifying the above equation by taking LCM on left hand side, then we will get
$\begin{aligned} & \sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ =\dfrac{\sqrt{3}\left( \sqrt{3}-2 \right)+2\left( 2 \right)}{2\sqrt{3}} \\\ & \Rightarrow \sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ =\dfrac{3-2\sqrt{3}+4}{2\sqrt{3}} \\\ & \Rightarrow \sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ =\dfrac{7-2\sqrt{3}}{2\sqrt{3}} \\\ \end{aligned}$
Adding the value of $\cot 90{}^\circ$ on both sides of the above equation, then we will have
$\begin{aligned} & \sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ +\cot 90{}^\circ =\dfrac{7-2\sqrt{3}}{2\sqrt{3}}+0 \\\ & \Rightarrow \sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ +\cot 90{}^\circ =\dfrac{7-2\sqrt{3}}{2\sqrt{3}} \\\ \end{aligned}$
Rationalizing the above value by dividing and multiplying with $\dfrac{\sqrt{3}}{\sqrt{3}}$ on left hand side, then we will get
$\begin{aligned} & \sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ +\cot 90{}^\circ =\dfrac{7-2\sqrt{3}}{2\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}} \\\ & \Rightarrow \sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ +\cot 90{}^\circ =\dfrac{\sqrt{3}\left( 7-2\sqrt{3} \right)}{2\sqrt{3}\times \sqrt{3}} \\\ & \Rightarrow \sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ +\cot 90{}^\circ =\dfrac{7\sqrt{3}-2\times 3}{2\times 3} \\\ & \Rightarrow \sin 0{}^\circ +\cos 30{}^\circ -\tan 45{}^\circ +\csc 60{}^\circ +\cot 90{}^\circ =\dfrac{-6+7\sqrt{3}}{6} \\\ \end{aligned}$
Hence option (B) is the correct answer.

Note: Here we have the trigonometric values for the given angles in the trigonometric table, so we have directly used them. Sometimes they won’t give the standard angles, then we need to calculate the trigonometric ratios for a given angle by using the half angle formulas or some other trigonometric formulas based on the given angle.