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

Hint: To find the value of $ \tan 27{}^\circ $ , we will use the fundamental trigonometric formulas such as $ \tan x=\dfrac{\sin x}{\cos x} $ , and we will also use the identity such as $ \sin \left( A-B \right)=\sin A\cos B-\sin B\cos A $ , $ \cos \left( A-B \right)=\cos A\cos B+\sin A\cos B $ . Also, we will use the value of $ \cos 18{}^\circ =\sin 72{}^\circ =\dfrac{1}{4}\left( \sqrt{5}-1 \right) $ and $ \cos 72{}^\circ =\sin 18{}^\circ =\dfrac{1}{4}\left( \sqrt{10+2\sqrt{5}} \right) $ .

Complete step by step answer:
We can see that the above-given question is of trigonometry, so we will use trigonometric identity to solve the above question.
Since we have to find the value of $ \tan 27{}^\circ $.
We can write $ \tan 27{}^\circ $ as $ \tan \left( 45{}^\circ -18{}^\circ \right) $ .
And, we know that $ \tan x=\dfrac{\sin x}{\cos x} $ , so we can write:
 $ \Rightarrow \tan 27{}^\circ $ = $ \tan \left( 45{}^\circ -18{}^\circ \right) $ = $ \dfrac{\sin \left( 45{}^\circ -18{}^\circ \right)}{\cos \left( 45{}^\circ -18{}^\circ \right)} $ .
Now, we also know that $ \sin \left( A-B \right)=\sin A\cos B-\sin B\cos A $ and $ \cos \left( A-B \right)=\cos A\cos B+\sin A\cos B $ .
 $ =\dfrac{\sin 45{}^\circ \cos 18{}^\circ -\sin 18{}^\circ \cos 45{}^\circ }{\cos 45{}^\circ \cos 18{}^\circ +\sin 45{}^\circ \sin 18{}^\circ } $
Now, we will put the value of $ \sin 45{}^\circ =\cos 45{}^\circ =\dfrac{1}{\sqrt{2}} $ , and $ \cos 18{}^\circ =\dfrac{1}{4}\left( \sqrt{5}-1 \right) $ , $ \sin 18{}^\circ =\dfrac{1}{4}\left( \sqrt{10+2\sqrt{5}} \right) $ in the above expression.
 $ \Rightarrow \dfrac{\dfrac{1}{\sqrt{2}}\times \dfrac{1}{4}\left( \sqrt{10+2\sqrt{5}} \right)-\dfrac{1}{4}\left( \sqrt{5}-1 \right)\times \dfrac{1}{\sqrt{2}}}{\dfrac{1}{\sqrt{2}}\times \dfrac{1}{4}\left( \sqrt{10+2\sqrt{5}} \right)+\dfrac{1}{\sqrt{2}}\times \dfrac{1}{4}\left( \sqrt{5}-1 \right)} $
 $ \Rightarrow \dfrac{\dfrac{1}{4\sqrt{2}}\left( \sqrt{10+2\sqrt{5}}-\sqrt{5}+1 \right)}{\dfrac{1}{4\sqrt{2}}\left( \sqrt{10+2\sqrt{5}}+\sqrt{5}-1 \right)} $
 $ \Rightarrow \dfrac{\left( \sqrt{10+2\sqrt{5}}-\sqrt{5}+1 \right)}{\left( \sqrt{10+2\sqrt{5}}+\sqrt{5}-1 \right)} $
Now, we will rationalize it above term to simplify it.
 $ \Rightarrow \dfrac{\left( \sqrt{10+2\sqrt{5}}-\sqrt{5}+1 \right)\left( \left( \sqrt{10+2\sqrt{5}}-\sqrt{5}+1 \right) \right)}{\left( \sqrt{10+2\sqrt{5}}+\sqrt{5}-1 \right)\left( \sqrt{10+2\sqrt{5}}-\sqrt{5}+1 \right)} $
 $ \Rightarrow \dfrac{{{\left( \sqrt{10+2\sqrt{5}}-\sqrt{5}+1 \right)}^{2}}}{{{\left( \sqrt{10+2\sqrt{5}} \right)}^{2}}-{{\left( \sqrt{5}-1 \right)}^{2}}} $
 $ \Rightarrow \dfrac{10+2\sqrt{5}+{{\left( \sqrt{5}-1 \right)}^{2}}-2\left( \sqrt{10+2\sqrt{5}} \right)\left( \sqrt{5}-1 \right)}{{{\left( \sqrt{10+2\sqrt{5}} \right)}^{2}}-{{\left( \sqrt{5}-1 \right)}^{2}}} $
 $ \Rightarrow \dfrac{10+2\sqrt{5}+\left( 6-2\sqrt{5} \right)-2\left( \sqrt{10+2\sqrt{5}} \right)\left( \sqrt{5}-1 \right)}{10+2\sqrt{5}-\left( 6-2\sqrt{5} \right)} $
 $ \Rightarrow \dfrac{16-2\left( \sqrt{10+2\sqrt{5}} \right)\left( \sqrt{5}-1 \right)}{4+4\sqrt{5}} $
 $ \Rightarrow \dfrac{16-2\left( \sqrt{10+2\sqrt{5}} \right)\left( \sqrt{5}-1 \right)}{4\left( 1+\sqrt{5} \right)} $
Now, we will again rationalize the above term with respect to $ \left( \sqrt{5}+1 \right) $ .
 $ \Rightarrow =\dfrac{\left( 16-2\left( \sqrt{10+2\sqrt{5}} \right)\left( \sqrt{5}-1 \right) \right)\left( \sqrt{5}-1 \right)}{4\left( \sqrt{5}+1 \right)\times \left( \sqrt{5}-1 \right)} $
 $ \Rightarrow \dfrac{\left( 16-2\left( \sqrt{10+2\sqrt{5}} \right)\left( \sqrt{5}-1 \right) \right)\left( \sqrt{5}-1 \right)}{4\times 4} $
 $ \Rightarrow \dfrac{\left( 16\left( \sqrt{5}-1 \right)-2\left( \sqrt{10+2\sqrt{5}} \right){{\left( \sqrt{5}-1 \right)}^{2}} \right)}{16} $
 $ \Rightarrow \dfrac{\left( 16\left( \sqrt{5}-1 \right)-2\left( \sqrt{10+2\sqrt{5}} \right)\left( 6-2\sqrt{5} \right) \right)}{16} $
 $ \Rightarrow \left( \sqrt{5}-1 \right)-\dfrac{\left( \left( \sqrt{10+2\sqrt{5}} \right)\left( 4-2\sqrt{5} \right) \right)}{8} $
Now, we will take 2 common from $ \left( 4-2\sqrt{5} \right) $ :
 $ \Rightarrow \left( \sqrt{5}-1 \right)-\dfrac{2\left( \left( \sqrt{10+2\sqrt{5}} \right)\left( 3-\sqrt{5} \right) \right)}{8} $
Now, we will take $ \left( 3-\sqrt{5} \right) $ in the square root to simplify it more:
 $ \Rightarrow \left( \sqrt{5}-1 \right)-\dfrac{\sqrt{\left( 10+2\sqrt{5} \right){{\left( 3-\sqrt{5} \right)}^{2}}}}{4} $
 $ \Rightarrow \left( \sqrt{5}-1 \right)-\dfrac{\sqrt{\left( 10+2\sqrt{5} \right)\left( 14-6\sqrt{5} \right)}}{4} $
 $ \Rightarrow \left( \sqrt{5}-1 \right)-\dfrac{\sqrt{\left( 140+28\sqrt{5}-60\sqrt{5}-60 \right)}}{4} $
 $ \Rightarrow \left( \sqrt{5}-1 \right)-\dfrac{\sqrt{\left( 80-32\sqrt{5} \right)}}{4} $
 $ \Rightarrow \left( \sqrt{5}-1 \right)-\dfrac{\sqrt{16\left( 5-2\sqrt{5} \right)}}{4} $
 $ \Rightarrow \left( \sqrt{5}-1 \right)-\dfrac{4\sqrt{\left( 5-2\sqrt{5} \right)}}{4} $
 $ \Rightarrow \left( \sqrt{5}-1 \right)-\sqrt{5-2\sqrt{5}} $
This is our required solution.

Note:
Students are required to note that when we are given $ \sec \theta $ , $ \operatorname{cosec}\theta $ , $ \tan \theta $ , and $ \cot \theta $ in the trigonometric expression then we always change them into $ \sin \theta $ and $ \cos \theta $ because we can easily simplify them. Students are also required to memorize the trigonometric formulas and values of special trigonometric values such as $ 15{}^\circ,18{}^\circ $, etc very well otherwise they will face problem while this try of question. Students are also requested to do calculations very carefully otherwise as we can see it is very lengthy and there are complicated terms.