Question

We also make use of the trigonometric relations on the values in cosine to secant. What can you say about $\cot {0^ \circ } = \dfrac{1}{{\tan {0^ \circ }}}$ . is it defined? Why? $\sec {0^ \circ } = 1$ Why?

Answer 1

Trigonometry is the branch of mathematics that deals with specific functions of angles and their application in calculations. Trigonometry is made up of two words trigon which means triangle and matron which means “to measure” ‘. There are six trigonometric ratios, “sine” abbreviated as “sin”, “cosine” abbreviated as “cos”, “tangent” as “tan”, “cotangent” as “cot”,” secant” as “sec” and “cosecant” as “cosec”. These trigonometric ratios are used for finding unknown values of angles and sides of a right-angled triangle.
Formula used:
These ratios have few relations among themselves and these are –
$\sin \theta = \dfrac{1}{\cos ec\theta }$
$\cos \theta = \dfrac{1}{{\sec \theta }}$
$\tan \theta = \dfrac{1}{{\cot \theta }}$=$\dfrac{{\sin \theta }}{{\cos \theta }}$
All these trigonometric ratios have different values at different angles and these values repeat after a particular angle.
The values of $\sin \theta$and $\cos \theta$at ${0^ \circ }$ are $0$ and $1$ respectively, i.e.,
$\sin {0^ \circ } = 0$and $\cos {0^ \circ } = 1$

Complete step by step answer:
We are aware that cot and tan are reciprocal of each other so,
$\cot {0^ \circ } = \dfrac{1}{{\tan {0^ \circ }}}$ also expressed as in the form of $\tan {0^ \circ } = \dfrac{1}{{\cot {0^ \circ }}}$
= $\dfrac{{\cos {0^ \circ }}}{{\sin {0^ \circ }}}$
On putting the values of $\cos {0^ \circ }$and $\sin {0^ \circ }$
=$\dfrac{1}{0}$
The form of such type is undermining form and we cannot define it. Thus $\cot {0^ \circ } = \dfrac{1}{{\tan {0^ \circ }}}$ is not defined.
We know that cos and sec are reciprocal functions of each other. Thus,
$\sec \theta = \dfrac{1}{{\cos \theta }}$
Now we shall substitute $\theta = 0^\circ$ in the above identity.
$\sec {0^ \circ } = \dfrac{1}{{\cos {0^ \circ }}}$
Since $\cos {0^ \circ } = 1$the above equation becomes as follows.
$\sec {0^ \circ } = \dfrac{1}{1}$
$\sec {0^ \circ } = 1$
Thus, we can say that $\cos {0^ \circ }$ is not defined while $\sec {0^ \circ }$ is $1$ .

Note:
Be attentive while placing the values of trigonometric ratios at different angles and don’t write incorrect values. Also, pay special attention while writing one trigonometric function in the form of others. Avoid confusion between different trigonometric ratios.