Answer
459.3k+ views
Hint:-In this question first we need to find the trigonometric relation between $\tan \theta {\text{ and cot}}\theta $.Then, calculate the value of $\tan {0^ \circ }$and put in the obtained relation to get the required value of $\cot {0^ \circ }$.
Complete step-by-step answer::
Using trigonometric relations, we know
$\cot \theta = \dfrac{1}{{\tan \theta }}$ eq.1
Since we know, $\tan {0^ \circ } = 0$ eq.2
And $\dfrac{1}{0} = {\text{ Not defined}}$ eq.3
Now put $\theta = {0^ \circ }$in eq.1,
We get
$ \Rightarrow \cot 0 = \dfrac{1}{{\tan 0}}$
Using eq.2 we can rewrite above equation as
$ \Rightarrow \cot 0 = \dfrac{1}{0}$
Now using eq.3 we can rewrite above equation as
$ \Rightarrow \cot 0 = {\text{ Not defined}}$
Hence, option D. is correct.
Note:- Whenever you get this type of question the key concept to solve this is to learn about mostly used trigonometric functions like $\sin \theta ,\cos \theta ,\tan \theta ,\sec \theta ,\cos {\text{ec}}\theta {\text{ and cot}}\theta {\text{ }}$.And their conversion property like too.
Complete step-by-step answer::
Using trigonometric relations, we know
$\cot \theta = \dfrac{1}{{\tan \theta }}$ eq.1
Since we know, $\tan {0^ \circ } = 0$ eq.2
And $\dfrac{1}{0} = {\text{ Not defined}}$ eq.3
Now put $\theta = {0^ \circ }$in eq.1,
We get
$ \Rightarrow \cot 0 = \dfrac{1}{{\tan 0}}$
Using eq.2 we can rewrite above equation as
$ \Rightarrow \cot 0 = \dfrac{1}{0}$
Now using eq.3 we can rewrite above equation as
$ \Rightarrow \cot 0 = {\text{ Not defined}}$
Hence, option D. is correct.
Note:- Whenever you get this type of question the key concept to solve this is to learn about mostly used trigonometric functions like $\sin \theta ,\cos \theta ,\tan \theta ,\sec \theta ,\cos {\text{ec}}\theta {\text{ and cot}}\theta {\text{ }}$.And their conversion property like too.
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