How do you evaluate $\csc \left( -22\pi/6 \right)-\cot \left( 19\pi/7 \right)+\sec \left( -17\pi/5 \right)$ ?

Hint: In this question we have been asked to evaluate the value of the given trigonometric expression $\csc \left( -22\pi/6 \right)-\cot \left( 19\pi/7 \right)+\sec \left( -17\pi/5 \right)$ . For doing that we will individually evaluate the value of each trigonometric function from our basic concepts and then add them.

Complete step by step solution:
Now considering from the question we have been asked to evaluate the value of the given trigonometric expression $\csc \left( -22\pi/6 \right)-\cot \left( 19\pi/7 \right)+\sec \left( -17\pi/5 \right)$ .
For doing that we will individually evaluate the value of each trigonometric function from our basic concepts and then add them.
We will first evaluate the value of \[\csc (-22\pi/6)=\csc \left( \dfrac{-22\pi }{6} \right)\] for that we will use the concept that gives the formula $\csc \theta =\dfrac{1}{\sin \theta }$ . So we will have $\Rightarrow \csc \left( \dfrac{-22\pi }{6} \right)=\dfrac{1}{\sin \left( \dfrac{-22\pi }{6} \right)}$ . We know that $\sin \left( -\theta \right)=-\sin \theta $ so we will have $\Rightarrow \dfrac{-1}{\sin \left( \dfrac{22\pi }{6} \right)}=\dfrac{-1}{\sin \left( \dfrac{11\pi }{3} \right)}$ . Now we will evaluate the value of $\sin \left( \dfrac{11\pi }{3} \right)$ which can be written as $\sin \left( 4\pi -\dfrac{\pi }{3} \right)$ . As we know that every trigonometric function repeats after a complete angle and the value of complete angle is given as $2\pi $ so we can write this as $ \sin \left( \dfrac{-\pi }{3} \right)=-\sin \left( \dfrac{\pi }{3} \right)$ . From the basic concepts we know that the value of $\sin \left( \dfrac{\pi }{3} \right)$ is $\dfrac{\sqrt{3}}{2}$ . Hence we have $\csc \left( -22\pi/6 \right)=\dfrac{2}{\sqrt{3}}$ .
Now we will evaluate the value of $\cot \left( 19\pi/7 \right)$ this can be written as $ \cot \left( 3\pi -\dfrac{2\pi }{7} \right)=\cot \left( \pi -\dfrac{2\pi }{7} \right)$ . This lies in the second quadrant, in that case cotangent function is negative so we will have $ -\cot \left( \dfrac{2\pi }{7} \right)$ . By using a calculator we will have $ \cot \left( \dfrac{19\pi }{7} \right)=-1.25$ .
Now we need to evaluate the value of $\sec (-17\pi /5)$ which can be written as $ \sec \left( \dfrac{-17\pi }{5} \right)=\dfrac{1}{\cos \left( \dfrac{17\pi }{5} \right)}=\dfrac{1}{\cos \left( 3\pi +\dfrac{2\pi }{5} \right)}$ because $\sec \left( -\theta \right)=\sec \theta $ and $\sec \theta =\dfrac{1}{\cos \theta }$ . Now we will have $ \dfrac{1}{\cos \left( \pi +\dfrac{2\pi }{5} \right)}=\dfrac{1}{-\cos \left( \dfrac{2\pi }{5} \right)}$ because it lies in third quadrant. By using a calculator we will have the value of $\cos \left( \dfrac{2\pi }{5} \right)=0.31$ . So we will have $\sec \left( \dfrac{-17\pi }{5} \right)=-3.24$
Hence now we will have
$\begin{align}
  & \csc \left( \dfrac{-22\pi }{6} \right)-\cot \left( \dfrac{19\pi }{7} \right)+\sec \left( \dfrac{-17\pi }{5} \right)=\dfrac{2}{\sqrt{3}}+1.25-3.24 \\
 & \Rightarrow 1.15-1.99=-0.84 \\
\end{align}$ .

Therefore we can conclude that the value of the given expression is $-0.84$ .

Note: While answering this question we should be sure with our concept and observations. If we observe the given expression correctly and apply our basic concepts then we will have an accurate result in a short span of time. We need to perform the calculations very carefully. We have four quadrants and eight trigonometric ratios. In the first quadrant, all trigonometric functions are positive. In the second quadrant, sine and cosecant trigonometric functions are positive and the remaining all are negative. In the third quadrant, tangent and cotangent trigonometric functions are positive and the remaining all are negative. In the fourth quadrant, cosecant and secant trigonometric functions are positive and remaining all are negative. Similarly we can derive the values of any trigonometric expressions for example if we consider the trigonometric expression \[\cos \left( \dfrac{-22\pi }{6} \right)\] which will result as $ \cos \left( \dfrac{-22\pi }{6} \right)=\cos \left( \dfrac{11\pi }{3} \right)\Rightarrow \cos \left( 4\pi -\dfrac{\pi }{3} \right)=\cos \left( \dfrac{\pi }{3} \right)\Rightarrow \dfrac{1}{2}$ .