Question

How do you evaluate $\cot {27^ \circ }$?

Answer 1

Hint: For solving this particular question we have to convert the given trigonometric function cotangent into its corresponding ratio that is tangent , then we have to put $\theta = {27^ \circ }$ , Then evaluate tangent of ${27^ \circ }$ , inverse of this will give the actual result.

Formula used: We used the trigonometric property i.e., $\cot \theta = \dfrac{1}{{\tan \theta }}$.
You must know that sin, cosine, and tangent are considered as major trigonometric functions,
hence we can derive the solutions for the equations com

Complete step by step solution:
We have to find the value of $\cot {27^ \circ }$ ,
For this we have to convert the given function as ,
$ \Rightarrow \cot \theta = \dfrac{1}{{\tan \theta }}$
Therefore, by putting $\theta = {27^ \circ }$ , we can write ,
$ \Rightarrow \cot {27^ \circ } = \dfrac{1}{{\tan {{27}^ \circ }}}$
After simplifying tangent of ${27^ \circ }$ ,
We will get the following result,
$ \Rightarrow \cot {27^ \circ } = \dfrac{1}{{0.51}}$
Now, we can write this as ,
$ \Rightarrow \cot {27^ \circ } \approx 1.96$
Hence, we get the required result.

Additional Information: Trigonometric equations can have two type of solutions one is principal
solution and other is general solution. Principal solution is the one where equation involves a
variable $0 \leqslant x \leqslant 2\pi $ . General solution is the one which involves an integer, say k and provides all the solutions of the trigonometric equation.

Note: You must know that sin, cosine, and tangent are considered as major trigonometric functions, hence we can derive the solutions for the equations comprising these trigonometric functions or ratios. We can also derive the solutions for the other three trigonometric functions such as secant,
cosecant, and cotangent with the help of the solutions which are already derived.