Question

The value of the following trigonometric expression
$\tan \left( \dfrac{\pi }{7} \right)+2\tan \left( \dfrac{2\pi }{7} \right)+4\tan \left( \dfrac{4\pi }{7} \right)+8\cot \left( \dfrac{8\pi }{7} \right)$ is equal to
(a).$\cos ec\left( \dfrac{2\pi }{7} \right)+\cot \left( \dfrac{2\pi }{7} \right) $
(b).\[\tan \left( \dfrac{\pi }{14} \right)-\cot \left( \dfrac{\pi }{14} \right)\]
(c).\[\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{1-\cos \left( \dfrac{2\pi }{7} \right)}\]
(d).\[\dfrac{1+\cos \left( \dfrac{\pi }{7} \right)+\cos \left( \dfrac{2\pi }{7} \right)}{\sin \left( \dfrac{\pi }{7} \right)+\sin \left( \dfrac{2\pi }{7} \right)}\]

Answer 1

Hint: First of all, this is a very tricky problem. If you blindly apply the formula of the sub-multiple angle of trigonometric expression you would not get the solution. So, see the following step very carefully. We will try to use the general method.

Complete step-by-step solution:
First, we will use the following trigonometric identity, which is slightly nontrivial, where A is any angle, which can have defined value of Tangent and Cotangent function.
\[\begin{align}
  & \cot \left( A \right)-\tan \left( A \right)=2\cot (2A) \\
 &\Rightarrow \tan \left( A \right)=\cot \left( A \right)-2\cot \left( 2A \right)\text{ }\!\![\!\!\text{ we will use this form }\!\!]\!\!\text{ } \\
 & \text{Here is the proof ,} \\
 & \text{cot}\left( A \right)-\tan \left( A \right) \\
 & =\dfrac{\cos A}{\sin A}-\dfrac{\sin A}{\cos A} \\
 & =\dfrac{{{\cos }^{2}}A-{{\sin }^{2}}A}{\sin A.\cos A}\text{ }..........................\text{(1)} \\
 &\text{Here we will multiply denominator and numerator of fraction (1) by 2 and use the relations : sin}\left( 2A \right)=2\sin (A)\cos (A) \\
 & \text{and cos (2A) = co}{{\text{s}}^{2}}(A)-si{{n}^{2}}(A)\text{ }\text{. } \\
 & =\dfrac{2\cos 2A}{\sin 2A} \\
 & =2\cot 2A
\end{align}\]
Here we will set $A=\dfrac{\pi }{7}$
Now see the problem; we will use the second form which had already mentioned earlier,
\[\tan \left( \dfrac{\pi }{7} \right)+2\tan \left( \dfrac{2\pi }{7} \right)+4\tan \left( \dfrac{4\pi }{7} \right)+8\cot \left( \dfrac{8\pi }{7} \right) \\
 =\cot \left( \dfrac{\pi }{7} \right)-2\cot \left( \dfrac{2\pi }{7} \right)+2\cot \left( \dfrac{2\pi }{7} \right)-4\cot \left( \dfrac{4\pi }{7} \right)+4\cot \left( \dfrac{4\pi }{7} \right)-8\cot \left( \dfrac{8\pi }{7} \right)+8\cot \left( \dfrac{8\pi }{7} \right) \\
=\cot \left( \dfrac{\pi }{7} \right)\text{ }\!\![\!\!\text{ after cancelling each term }\!\!]\!\!\text{ } \\
=\dfrac{\cos \left( \dfrac{\pi }{7} \right)}{\sin \left( \dfrac{\pi }{7} \right)}\text{ }......................................\text{(2)} \]
We will multiply both denominator and numerator of (2) by \[2\sin \left( \dfrac{\pi }{7} \right)\]
\[=\dfrac{2\cos \left( \dfrac{\pi }{7} \right)\sin \left( \dfrac{\pi }{7} \right)}{2{{\sin }^{2}}\left( \dfrac{\pi }{7} \right)}\]
Here we will use the following two formulae of sine and cosine function,
$\begin{align}
  & \sin 2\theta =2\sin \theta \cos \theta \\
 & 2si{{n}^{2}}\theta =1-\cos 2\theta \\
\end{align}$
After using the formula and put \[\theta =\dfrac{\pi }{7}\]
\[=\dfrac{\sin \left( \dfrac{2\pi }{7} \right)}{1-\cos \left( \dfrac{2\pi }{7} \right)}\]
Hence, option (c) is correct.
We are done.

Note: The first thing we have to keep in mind that blind guess would not work. If we use the formula of tangent and cotangent then we can not get the answer. At first, without previous knowledge, it’s very hard to predict the approach, but with some experience, you can able to solve the problem with this technique.