Question

How do you evaluate the value of six trigonometric functions given \[t = \pi \]?

Answer 1

Hint: We are given with the angle of which the six trigonometric functions are to be obtained. This angle is in radians but it is equal to \[{180^ \circ }\] in degrees. Now we can find the six functions with the help of two functions only. Those are sin and cos. Remaining four functions can be obtained with the help of these two only.

Complete step by step solution:
Given the angle is \[t = \pi \]
Now we know that,
\[\sin \pi = 0\]
This is the first function.
\[\cos \pi = - 1\]
Now the next function and their values can be obtained with the help of these two.
\[\tan \pi = \dfrac{{\sin \pi }}{{\cos \pi }}\]
We know that tan function is the ratio of sin to cos. Ow putting the values we can write,
\[\tan \pi = \dfrac{0}{{ - 1}} = 0\]
Now using the reciprocal method we can find the rest three functions.
\[\cot \pi = \dfrac{1}{{\tan \pi }} = \dfrac{1}{0} = ND\] This function is not defined for this angle.
\[\sec \pi = \dfrac{1}{{\cos \pi }} = \dfrac{1}{{ - 1}} = - 1\]
Last function is, \[\cos ec \pi = \dfrac{1}{{\sin \pi }} = \dfrac{1}{0} = ND\]

Note: Note that the values are just to be found. We can also note whether the values so obtained and signs of the respective functions in the quadrant are the same or not. Also note that the value with 0 by any number is 0. But the value having 0 in the denominator will be not defined.