Question

Find the value of the given trigonometric ratio, $\tan 15{}^{\circ }$ .

Answer 1

Hint: Use the formula of $\tan 2A$ along with the value of $\tan 30{}^{\circ }$ , to get a quadratic equation. Solve the quadratic equation to reach the required answer.

Complete step-by-step answer:
We know;
$\tan 30{}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}}$

The other commonly used trigonometric values include:
$\tan 0{}^{\circ }=0$
$\tan 45{}^{\circ }=1$
$\tan 60{}^{\circ }=\sqrt{3}$

Also, we have, the formula: $\tan 2A={\displaystyle \frac{2\tan A}{1-{\tan }^{2}A}}$

So, in the above formula substituting $A=15{}^{\circ }$ .
 $\therefore \tan 2A={\displaystyle \frac{2\tan A}{1-{\tan }^{2}A}}$
$\Rightarrow \tan (2\times 15{}^{\circ })={\displaystyle \frac{2\tan 15{}^{\circ }}{1-{\tan }^{2}15{}^{\circ }}}$
$\Rightarrow \tan 30{}^{\circ }={\displaystyle \frac{2\tan 15{}^{\circ }}{1-{\tan }^{2}15{}^{\circ }}}$

Putting the value of $\tan 30{}^{\circ }$ in the equation, we get;
${\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{2\tan 15{}^{\circ }}{1-{\tan }^{2}15{}^{\circ }}}$

On cross-multiplication, we get;
$1-{\tan }^{2}15{}^{\circ }=2\sqrt{3}\tan 15{}^{\circ }$
$\Rightarrow {\tan }^{2}15{}^{\circ }+2\sqrt{3}\tan 15{}^{\circ }-1=0$

So, the equation we get is a quadratic equation, and one of the roots of this quadratic equation would be the value of $\tan 15{}^{\circ }$.

We know, for a quadratic equation of the form $a{x}^2+bx+c=0$ .
$x={\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

Applying the formula to our quadratic equation, we have;
$\tan 15{}^{\circ }={\displaystyle \frac{-2\sqrt{3}\pm \sqrt{{\left(2\sqrt{3}\right)}^2-4\times 1\times (-1)}}{2\times 1}}$
$\Rightarrow \tan 15{}^{\circ }={\displaystyle \frac{-2\sqrt{3}\pm \sqrt{12+4}}{2}}$
$\Rightarrow \tan 15{}^{\circ }={\displaystyle \frac{-2\sqrt{3}\pm \sqrt{16}}{2}}$
$\Rightarrow \tan 15{}^{\circ }={\displaystyle \frac{-2\sqrt{3}\pm 4}{2}}$

We know, $15{}^{\circ }$ lies in the first quadrant.

According to the graph of $\tan (x)$ :

 $\tan (x)$ is positive when x lies in the first quadrant.

Therefore, $\tan 15{}^{\circ }$ is also positive.
$\therefore \tan 15{}^{\circ }={\displaystyle \frac{-2\sqrt{3}+4}{2}}$
$\Rightarrow \tan 15{}^{\circ }={\displaystyle \frac{(-\sqrt{3}+2)}{}}$
$\therefore \tan 15{}^{\circ }=2-\sqrt{3}$

Hence, the value of $\tan 15{}^{\circ }$ is $2-\sqrt{3}$ .

Note: Other useful formulas include:
$\tan (A+B)={\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B}}$
$\tan (A-B)={\displaystyle \frac{\tan A-\tan B}{1+\tan A\tan B}}$

And you are free to use any formula, just substitute the angles according to the need to get the desired values.

We can also find the value of $\tan 15{}^{\circ }$ using formula: $\tan (A-B)={\displaystyle \frac{\tan A-\tan B}{1+\tan A\tan B}}$ .

On Substituting A and B in the above formula, we get;
$A=45{}^{\circ }$
$B=30{}^{\circ }$

The equation becomes:
$\tan (A-B)={\displaystyle \frac{\tan A-\tan B}{1+\tan A\tan B}}$
$\Rightarrow \tan (45{}^{\circ }-30{}^{\circ })={\displaystyle \frac{\tan 45{}^{\circ }-\tan 30{}^{\circ }}{1+\tan 45{}^{\circ }\tan 30{}^{\circ }}}$
$$\Rightarrow \tan 15{}^{\circ }={\displaystyle \frac{1-\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}{1+1\times {\displaystyle \frac{1}{\sqrt{3}}}}}$$

Point to remember: whenever you try to find the value of $\sin 15{}^{\circ }$ , don’t use the formula of $\sin 2A$ , instead, go for the formula: $\cos 2A=1-2{\sin }^{2}A$ . The reason being, whenever you use the formula of $\sin 2A$ , you get both $\cos A$ and $\sin A$ to be unknown, making it difficult to solve.