Question

How do you find the value of $\cot {240^ \circ }$ using double angle or half angle identity ?

Answer 1

Here we are given with an angle with its trigonometric function. Actually ${240^ \circ }$ is a standard angle but we need to convert this angle in the basic angle with the help of double angle or half angle identity. Then we will use the angle with tangent function to find the value of $\cot {240^ \circ }$.

Complete step by step solution:
Given that $\cot {240^ \circ }$ is the angle whose value is to be found.
We know that $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
So ${240^ \circ } = 2 \times {120^ \circ }$
So we can write
$\tan {240^ \circ } = \dfrac{{2\tan {{120}^ \circ }}}{{1 - {{\tan }^2}{{120}^ \circ }}}$
Also we can write ${120^ \circ } = 2 \times {60^ \circ }$
So the identity will be,
$\tan {120^ \circ } = \dfrac{{2\tan {{60}^ \circ }}}{{1 - {{\tan }^2}{{60}^ \circ }}}$
We know that $\tan {60^ \circ } = \sqrt 3$

Putting this value in the identity we get,
$\tan {120^ \circ } = \dfrac{{2\sqrt 3 }}{{1 - {{\left( {\sqrt 3 } \right)}^2}}}$
On calculating the square we get,
$\tan {120^ \circ } = \dfrac{{2\sqrt 3 }}{{1 - 3}}$
$\Rightarrow\tan {120^ \circ } = \dfrac{{2\sqrt 3 }}{{ - 2}}$
On cancelling 2 we get,
$\tan {120^ \circ } = - \sqrt 3$
Now we will put this value in the identity of ${120^ \circ }$
$\tan {240^ \circ } = \dfrac{{2\left( { - \sqrt 3 } \right)}}{{1 - {{\left( { - \sqrt 3 } \right)}^2}}}$

Taking the square,
$\tan {240^ \circ } = \dfrac{{2\left( { - \sqrt 3 } \right)}}{{1 - 3}}$
$\Rightarrow\tan {240^ \circ } = \dfrac{{ - 2\left( {\sqrt 3 } \right)}}{{ - 2}}$
Cancelling -2,
$\tan {240^ \circ } = \sqrt 3$
Now we know that
$\cot {240^ \circ } = \dfrac{1}{{\tan {{240}^ \circ }}}$
So putting the value,
$\therefore\cot {240^ \circ } = \dfrac{1}{{\sqrt 3 }}$

Hence, the value of $\cot {240^ \circ }$ is $\dfrac{1}{{\sqrt 3 }}$.

Note: We used half angle identity with double angle form. Also note that if we cannot reach the trigonometric function asked we need to take help of other such trigonometric functions that include or may help to reach the trigonometric function. We can go with the term like we know the value of $n\pi$ form for the same trigonometric function.