Question

Find the value of $ \tan {{15}^{\circ }} $ ?
(a) $ 2+\sqrt{3} $
(b) $ 2-\sqrt{3} $
(c) $ \sqrt{3}-2 $
(d) $ \dfrac{1}{2-\sqrt{5}} $

Answer 1

Hint: We started solving the problem by writing $ \tan {{15}^{\circ }} $ as $ \tan \left( {{45}^{\circ }}-{{30}^{\circ }} \right) $ . We then make use of the fact that $ \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\left( \tan A\times \tan B \right)} $ , $ \tan {{45}^{\circ }}=1 $ and $ \tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}} $ to proceed through the problem. We then multiply the numerator and denominator with the conjugate rational surd of denominator to proceed further through the problem. We then make the necessary calculations to get the required value of $ \tan {{15}^{\circ }} $ .

Complete step by step answer:
According to the problem, we are asked to find the value of $ \tan {{15}^{\circ }} $ .
We know that $ \tan {{15}^{\circ }}=\tan \left( {{45}^{\circ }}-{{30}^{\circ }} \right) $ ---(1).
We can see that $ \tan \left( {{45}^{\circ }}-{{30}^{\circ }} \right) $ resembles $ \tan \left( A-B \right) $ . We know that $ \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\left( \tan A\times \tan B \right)} $ . Let us use this result in equation (1).
So, we get $ \tan {{15}^{\circ }}=\dfrac{\tan {{45}^{\circ }}-\tan {{30}^{\circ }}}{1+\left( \tan {{45}^{\circ }}\times \tan {{30}^{\circ }} \right)} $ ---(2).
We know that $ \tan {{45}^{\circ }}=1 $ and $ \tan {{30}^{\circ }}=\dfrac{1}{\sqrt{3}} $ . Let us substitute this results in equation (2).
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+\left( 1\times \dfrac{1}{\sqrt{3}} \right)} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{\dfrac{\sqrt{3}-1}{\sqrt{3}}}{1+\dfrac{1}{\sqrt{3}}} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{\dfrac{\sqrt{3}-1}{\sqrt{3}}}{\dfrac{\sqrt{3}+1}{\sqrt{3}}} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1} $ ---(3).
Let us multiply the numerator and denominator of equation (3) with $ \sqrt{3}-1 $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{\left( \sqrt{3}+1 \right)\times \left( \sqrt{3}-1 \right)} $ ---(4).
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ and $ \left( a-b \right)\times \left( a+b \right)={{a}^{2}}-{{b}^{2}} $ . Let us use these results in equation (4).
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{{{\left( \sqrt{3} \right)}^{2}}+{{\left( 1 \right)}^{2}}-2\left( \sqrt{3} \right)\left( 1 \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( 1 \right)}^{2}}} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{3+1-2\sqrt{3}}{3-1} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{4-2\sqrt{3}}{2} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=2-\sqrt{3} $ .
So, we have found the value of $ \tan {{15}^{\circ }} $ as $ 2-\sqrt{3} $ .
 $\therefore$ The correct option for the given problem is (b).

Note:
We can see that the given problem contains a huge amount of calculation, so we need to perform each step carefully in order to avoid confusion and calculation mistakes. We can also solve the given problem as shown below:
We know that $ \tan {{15}^{\circ }}=\tan \left( {{60}^{\circ }}-{{45}^{\circ }} \right) $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{\tan {{60}^{\circ }}-\tan {{45}^{\circ }}}{1+\left( \tan {{60}^{\circ }}\times \tan {{45}^{\circ }} \right)} $ ---(2).
We know that $ \tan {{45}^{\circ }}=1 $ and $ \tan {{60}^{\circ }}=\sqrt{3} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{1+\left( \sqrt{3}\times 1 \right)} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{1+\sqrt{3}} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1} $ .
Let us multiply the numerator and denominator with $ \sqrt{3}-1 $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{{{\left( \sqrt{3}-1 \right)}^{2}}}{\left( \sqrt{3}+1 \right)\times \left( \sqrt{3}-1 \right)} $ .
We know that $ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ and $ \left( a-b \right)\times \left( a+b \right)={{a}^{2}}-{{b}^{2}} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{{{\left( \sqrt{3} \right)}^{2}}+{{\left( 1 \right)}^{2}}-2\left( \sqrt{3} \right)\left( 1 \right)}{{{\left( \sqrt{3} \right)}^{2}}-{{\left( 1 \right)}^{2}}} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{3+1-2\sqrt{3}}{3-1} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=\dfrac{4-2\sqrt{3}}{2} $ .
 $ \Rightarrow \tan {{15}^{\circ }}=2-\sqrt{3} $ .
So, we have found the value of $ \tan {{15}^{\circ }} $ as $ 2-\sqrt{3} $ .