Question

The value of $\tan \dfrac{\pi }{5}+2\tan \dfrac{2\pi }{5}+4\cot \dfrac{4\pi }{5}$ is:
a). $\cot \dfrac{\pi }{5}$
b). $\cot \dfrac{2\pi }{5}$
c). $\cot \dfrac{4\pi }{5}$
d). $\cos \dfrac{3\pi }{5}$

Answer 1

To solve the above question we will replace tangent function by the ratio of the sine to the cosine function and cot by the ratio of cosine to sine function. Then, we will us the trigonometric properties like $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta$, $\sin 2\theta =2\sin \theta \cos \theta$etc. and then solve them to get the simplified form as given in the options.

Complete step by step answer:
Since, we have to find the value of $\tan \dfrac{\pi }{5}+2\tan \dfrac{2\pi }{5}+4\cot \dfrac{4\pi }{5}$.
We will start solving the above question just by replacing tangent function by the ratio of the sine to the cosine function and cot by the ratio of cosine to sine function (i.e. $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$) .
So, we will write $\tan \dfrac{\pi }{5}+2\tan \dfrac{2\pi }{5}+4\cot \dfrac{4\pi }{5}$ as the function of sine and cosine function.
Let’s take it as A.
$A=\dfrac{\sin \dfrac{\pi }{5}}{\cos \dfrac{\pi }{5}}+2\dfrac{\sin \dfrac{2\pi }{5}}{\cos \dfrac{2\pi }{5}}+4\dfrac{\cos \dfrac{4\pi }{5}}{\sin \dfrac{4\pi }{5}}$
Now, we will write use the property $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta$ and $\sin 2\theta =2\sin \theta \cos \theta$
We will write $\cos \dfrac{4\pi }{5}$ as ${{\cos }^{2}}\dfrac{2\pi }{5}-{{\sin }^{2}}\dfrac{2\pi }{5}$ and $\sin \dfrac{4\pi }{5}$ as $2\sin \dfrac{2\pi }{5}\cos \dfrac{2\pi }{5}$ , then we will get:
$\Rightarrow A=\dfrac{\sin \dfrac{\pi }{5}}{\cos \dfrac{\pi }{5}}+2\dfrac{\sin \dfrac{2\pi }{5}}{\cos \dfrac{2\pi }{5}}+4\times \dfrac{\left( {{\cos }^{2}}\dfrac{2\pi }{5}-{{\sin }^{2}}\dfrac{2\pi }{5} \right)}{2\sin \dfrac{2\pi }{5}\cos \dfrac{2\pi }{5}}$
$\Rightarrow A=\dfrac{\sin \dfrac{\pi }{5}}{\cos \dfrac{\pi }{5}}+2\dfrac{\sin \dfrac{2\pi }{5}}{\cos \dfrac{2\pi }{5}}+2\dfrac{\left( {{\cos }^{2}}\dfrac{2\pi }{5}-{{\sin }^{2}}\dfrac{2\pi }{5} \right)}{\sin \dfrac{2\pi }{5}\cos \dfrac{2\pi }{5}}$
Now, after take LCM we will get:
$\Rightarrow A=\dfrac{\sin \dfrac{\pi }{5}}{\cos \dfrac{\pi }{5}}+\dfrac{2{{\sin }^{2}}\dfrac{2\pi }{5}+2{{\cos }^{2}}\dfrac{2\pi }{5}-2{{\sin }^{2}}\dfrac{2\pi }{5}}{\cos \dfrac{2\pi }{5}\sin \dfrac{2\pi }{5}}$
$\Rightarrow A=\dfrac{\sin \dfrac{\pi }{5}}{\cos \dfrac{\pi }{5}}+\dfrac{2{{\cos }^{2}}\dfrac{2\pi }{5}}{\cos \dfrac{2\pi }{5}\sin \dfrac{2\pi }{5}}$
$\Rightarrow A=\dfrac{\sin \dfrac{\pi }{5}}{\cos \dfrac{\pi }{5}}+\dfrac{2\cos \dfrac{2\pi }{5}}{\sin \dfrac{2\pi }{5}}$
Now, we will again use the property $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta$ and $\sin 2\theta =2\sin \theta \cos \theta$ , to write $\cos \dfrac{2\pi }{5}$ as ${{\cos }^{2}}\dfrac{\pi }{5}-{{\sin }^{2}}\dfrac{\pi }{5}$ and $\sin \dfrac{2\pi }{5}$ as $2\sin \dfrac{\pi }{5}\cos \dfrac{\pi }{5}$.
$\Rightarrow A=\dfrac{\sin \dfrac{\pi }{5}}{\cos \dfrac{\pi }{5}}+\dfrac{2\left( {{\cos }^{2}}\dfrac{\pi }{5}-{{\sin }^{2}}\dfrac{\pi }{5} \right)}{2\cos \dfrac{\pi }{5}\sin \dfrac{\pi }{5}}$
Now, again after taking LCM, we will get:
$\Rightarrow A=\dfrac{{{\sin }^{2}}\dfrac{\pi }{5}+{{\cos }^{2}}\dfrac{\pi }{5}-{{\sin }^{2}}\dfrac{\pi }{5}}{\cos \dfrac{\pi }{5}\sin \dfrac{\pi }{5}}$
$\Rightarrow A=\dfrac{{{\cos }^{2}}\dfrac{\pi }{5}}{\cos \dfrac{\pi }{5}\sin \dfrac{\pi }{5}}$
After cancelling $\cos \dfrac{\pi }{5}$ from numerator and denominator we will get:
$\Rightarrow A=\dfrac{\cos \dfrac{\pi }{5}}{\sin \dfrac{\pi }{5}}$
Since, we know that $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$, so we can write $\dfrac{\cos \dfrac{\pi }{5}}{\sin \dfrac{\pi }{5}}$ as $\cot \dfrac{\pi }{5}$ .

So, the correct answer is “Option b”.

Note: Students are required to memorize all the formula and whenever they have been given a question which contains tangent functions, they should convert it into a sine and cosine function. It will always make our calculation easier and we can easily solve them by using different properties of trigonometry.