Question

Value of $\text{tan15}{}^{\circ }$ (this question has multiple correct option)
A.$\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$
B.$2-\sqrt{3}$
C. $2+\sqrt{3}$
D.$\sqrt{3}-1$

Answer 1

Hint: To calculate the value of $\text{tan15}{}^{\circ }$, we can use $\text{tan(A-B)}$ and we can also use $\text{tan2A}$.

Complete step by step answer:
We have $\text{tan15}{}^{\circ }$
WE can write it as,
$\Rightarrow \text{tan15}{}^{\circ }=\text{tan(}{{45}^{\circ }}-{{30}^{\circ }}\text{)}$
$\Rightarrow \text{tan(1}{{5}^{\circ }}\text{) = }\dfrac{\text{tan4}{{\text{5}}^{\circ }}-\text{tan3}{{\text{0}}^{\circ }}}{1-\text{tan4}{{\text{5}}^{\circ }}\times \text{tan3}{{\text{0}}^{\circ }}}$ $\left\{ \because \text{tan(A-B)}=\text{ }\dfrac{\text{tanA - tanB}}{1+\text{tanA}\times \text{tanB}} \right\}$
  We know that , $\text{tan4}{{\text{5}}^{\circ }}=1$ and $\text{tan3}{{\text{0}}^{\circ }}=\dfrac{1}{\sqrt{3}}$
Hence, $\Rightarrow \text{tan(1}{{5}^{\circ }}\text{) = }\dfrac{1-\dfrac{1}{\sqrt{3}}}{1+\dfrac{1}{\sqrt{3}}}$
               $\Rightarrow \text{tan(1}{{5}^{\circ }}\text{)}=\dfrac{\dfrac{\sqrt{3}-1}{\sqrt{3}}}{\dfrac{\sqrt{3}+1}{\sqrt{3}}}$
$\Rightarrow \text{tan(1}{{5}^{\circ }}\text{)}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$
It is option A.
Now we can also write $\text{tan15}{}^{\circ }$as,
$\text{tan(2}\times \text{15}{}^{\circ })=\text{tan}({{30}^{\circ }})$
$\text{tan(2}\times \text{15}{}^{\circ })=\text{tan(2A)}$
 And, $\text{tan(2A) = }\dfrac{2\text{tanA}}{1-\text{ta}{{\text{n}}^{2}}\text{A}}$
 Also, $\text{tan3}{{\text{0}}^{\circ }}=\dfrac{1}{\sqrt{3}}$
Therefore,
$\text{tan(2}\times \text{15}{}^{\circ })=\text{tan3}{{\text{0}}^{\circ }}$
  $\Rightarrow \dfrac{2\text{tan1}{{\text{5}}^{\circ }}}{1-\text{ta}{{\text{n}}^{2}}{{15}^{\circ }}}=\dfrac{1}{\sqrt{3}}$
   $\Rightarrow 2\text{tan(}{{15}^{\circ }})\times \sqrt{3}=1-\text{ta}{{\text{n}}^{2}}{{15}^{\circ }}$
   $\Rightarrow \text{ta}{{\text{n}}^{2}}{{15}^{\circ }}+2\sqrt{3}\text{tan1}{{\text{5}}^{\circ }}-1=0$
Now we can let $x=\tan ({{15}^{\circ }})$
Hence equation is
$\Rightarrow {{x}^{2}}+2\sqrt{3}x-1=0$
It is a quadratic equation of the form $a{{x}^{2}}+bx+c=0$ We can solve it by quadratic formula.
On comparing with $a{{x}^{2}}+bx+c=0$
$a=1\,,b=2\sqrt{3},\,c\,=-1$
By quadratic formula we can write
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$\Rightarrow x=\dfrac{-2\sqrt{3}\pm \sqrt{{{(2\sqrt{3})}^{2}}-4\times 1\times -1}}{2\times 1}$
$\Rightarrow x=\dfrac{-2\sqrt{3}\pm \sqrt{12+4}}{2}$
$\Rightarrow x=\dfrac{-2\sqrt{3}\pm \sqrt{16}}{2}$
$\Rightarrow x=\dfrac{-2\sqrt{3}\pm 4}{2}$
$\Rightarrow x=-\sqrt{3}\pm 2$
On substituting $x=\tan ({{15}^{\circ }})$
$\tan ({{15}^{\circ }})=-\sqrt{3}+2\,or\,\tan ({{15}^{\circ }})=-\sqrt{3}-2\,$
We know $\text{tan1}{{\text{5}}^{\circ }}$fall in 1st quadrant
Therefore, $\text{tan1}{{\text{5}}^{\circ }} > 0$
Hence, $\text{tan1}{{\text{5}}^{\circ }}=2-\sqrt{3}$

So, the correct answers are “Option A and B”.

Note: As in trigonometric we have value of general angle as ${{0}^{\circ }},\,{{30}^{\circ }}\,,\,{{45}^{\circ }},\,{{60}^{\circ }},\,{{90}^{\circ }}$ which values we know from trigonometry table. As in given question we know $\text{tan1}{{\text{5}}^{\circ }}$can’t be calculated directly then we have to transform given statement like $\text{tan1}{{\text{5}}^{\circ }}$in expressing we can conclude like $\text{tan(4}{{\text{5}}^{\circ }}\text{-3}{{\text{0}}^{\circ }})$and $\text{tan(2}\times \text{1}{{\text{5}}^{\circ }})$we did in this question.