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Question: How you evaluate \(\cot \left( { - \dfrac{\pi }{6}} \right)?\)...

How you evaluate cot⁡(−π6)?\cot \left( { - \dfrac{\pi }{6}} \right)?

Explanation

Solution

In this question, to find the cotangent of negative argument or angle, i.e. cot⁡(−π6)\cot \left( { - \dfrac{\pi }{6}} \right), we have to use the formula for negative angles for tangent and cotangent, that is given as

tan⁡(−x)=−tan⁡x cot⁡(−x)=−cot⁡x  \tan ( - x) = - \tan x \\\ \cot ( - x) = - \cot x \\\
Use this formula to convert the negative argument or angle into positive one for cotangent.

Complete step by step solution:
In this question, we have to find the value of cot⁡(−π6)\cot \left( { - \dfrac{\pi }{6}} \right)

As we know that cotangent and tangent of an argument becomes negative with equal magnitude when same argument also becomes negative, this can be understood as follows

Let us take value of tan⁡x=y\tan x = y, then the value of tangent when its argument that is xx becomes negative, can be given as
tan⁡(−x)=−tan⁡x\tan ( - x) = - \tan x

And we know that tan⁡x=y\tan x = y,
∴tan⁡(−x)=−y\therefore \tan ( - x) = - y

Using this method in order to find the value of cot⁡(−π6)\cot \left( { - \dfrac{\pi }{6}} \right)
⇒cot⁡(−π6)=−cot⁡π6\Rightarrow \cot \left( { - \dfrac{\pi }{6}} \right) = - \cot \dfrac{\pi }{6}

Now we have to put the value of cot⁡π6\cot \dfrac{\pi }{6} above in order to get the desired answer,

As we know that cot⁡π6=3\cot \dfrac{\pi }{6} = \sqrt 3

Putting this value above, we will get
⇒cot⁡(−π6)=−cot⁡π6=−3\Rightarrow \cot \left( { - \dfrac{\pi }{6}} \right) = - \cot \dfrac{\pi }{6} = - \sqrt 3

Therefore we got the value of cot⁡(−π6)=−3\cot \left( { - \dfrac{\pi }{6}} \right) = - \sqrt 3 by evaluating it.

Note: Generally students do not remember the general values of secant, cosecant, and cotangent, but if you have remembered the general values of sine, cosine and tangent then you don’t have to worry about it, since sine and cosine are multiplicative inverse of each other similar case with cosine and secant, and tangent and cotangent.

So you can write their values as follows csc⁡x=1sin⁡x,  sec⁡x=1cos⁡x  and  cot⁡x=1tan⁡x\csc x = \dfrac{1}{{\sin x}},\;\sec x = \dfrac{1}{{\cos x}}\;{\text{and}}\;\cot x = \dfrac{1}{{\tan x}}

You can also solve this question by adding 2π2\pi to the argument and then finding the
content of the argument that comes after addition. Try this by yourself and check the answer.