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Find the value of the given trigonometric identity whose angle is given, the identity is \[\cos ec(210)\] ?

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Last updated date: 04th Mar 2024
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IVSAT 2024
Answer
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Hint: To find the exact known value of trigonometric function of larger angels we have to break the angle by using the property such that the function did not change its property. For bigger angles we can’t learn the direct values hence it needs to be broken down into smaller angles. Also we know that “sin” function does not change its property in the first and second quadrant.

Formulae Used:
\[\sin (180 + \theta ) = - \sin \theta \]

Complete step by step answer:
The given function is \[\cos ec(210)\]
We know that “cosec” function is reversible of “sin” function, lets use this property:
We get:
\[\cos ec(210) = \dfrac{1}{{\sin (210)}}\]
Since the behavior of “sin” function is known to us for different coordinates.
Now we have to find out the behavior of “sin” function in third coordinate, on solving we get:
\[\sin (180 + \theta ) = \sin (90 + (90 + \theta )) = \cos (90 + \theta ) = - \sin \theta (\cos \,behaves\,as\,negative\,function\,in\ sec ond\,quadrant) \\
\Rightarrow \sin (180 + \theta ) = - \sin \theta \\ \]
From here we can see the behavior of “sin” function in third quadrant, now using this in our question we get:
\[\sin (210) = \sin (180 + 30) \\
\Rightarrow\sin (210) = - \sin (30) \\
\Rightarrow\sin (210) = - \dfrac{1}{2}\]
Now for finding the value of the “cosec” function we can reciprocate the answer, on solving we get:
\[\cos ec(210) = \dfrac{1}{{\sin (210)}} \\
\Rightarrow\cos ec(210) = \dfrac{1}{{ - \dfrac{1}{2}}} \\
\therefore\cos ec(210) = - 2\]
This is our final answer for the given question.

Note:Trigonometric values for the functions are known for smaller angles, because it is impossible to learn the value of every angle hence some basic small angle like zero, thirty, forty five, sixty, and ninety degrees are known, for solving with bigger angles you have to rearrange the angle by using the properties to get in between these angles.
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