Question

The value of $\tan \dfrac{\pi }{8}$ is equal to ?
A) $\dfrac{1}{2}$
B) $\sqrt 2 - 1$
C) $\dfrac{1}{{\sqrt 2 + 1}}$
D) $1 - \sqrt 2 $

Answer 1

Hint:
We can consider $\dfrac{\pi }{8}$ as half of the angle $\dfrac{\pi }{4}$. Then we can apply the result of $\tan 2A$. Substituting the known values we get a quadratic equation in $\tan \dfrac{\pi }{8}$. Solving it we get the answer.

Useful formula:
The standard form of the second degree equation is $a{x^2} + bx + c = 0$.
And the solution of such an equation is given by,
Also we have the trigonometric formula:
$\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$

Complete step by step solution:
We are asked to find $\tan \dfrac{\pi }{8}$.
We can write it as $\tan \dfrac{\pi }{{4 \times 2}} = \tan \dfrac{{\dfrac{\pi }{4}}}{2}$
We know that $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
 Let,
$A = \dfrac{\pi }{8} \Rightarrow 2A = \dfrac{\pi }{4}$
Substituting we get,
$\tan \dfrac{\pi }{4} = \dfrac{{2\tan \dfrac{\pi }{8}}}{{1 - {{\tan }^2}\dfrac{\pi }{8}}}$
We know that ,
$\tan \dfrac{\pi }{4} = 1$
Substituting this we get,
$1 = \dfrac{{2\tan \dfrac{\pi }{8}}}{{1 - {{\tan }^2}\dfrac{\pi }{8}}}$
Again let
$\tan \dfrac{\pi }{8} = x$
This gives,
$1 = \dfrac{{2x}}{{1 - {x^2}}}$
Cross-multiplying we get,
$1 - {x^2} = 2x$
Rearranging we get,
$x^2 + 2x -1 =0$
This can be compared with the standard form of second degree equation $a{x^2} + bx + c = 0$.
Here, $a = 1,b = 2,c = - 1$
And the solution of such an equation is given by,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So here we have,
$x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times - 1} }}{{2 \times 1}}$
Simplifying we get,
$x = \dfrac{{ - 2 \pm \sqrt {4 + 4} }}{2}$
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt 8 }}{2}$
Since $\sqrt 8 = \sqrt {4 \times 2} = 2\sqrt 2 $, we have
$ \Rightarrow x = \dfrac{{ - 2 \pm 2\sqrt 2 }}{2}$
Cancelling $2$ from numerator and denominator we have,
$ \Rightarrow x = - 1 \pm \sqrt 2 $
Substituting back for $x$ we have,
$\tan \dfrac{\pi }{8} = - 1 \pm \sqrt 2 $
That is, $\tan \dfrac{\pi }{8} = - 1 + \sqrt 2 $ or $\tan \dfrac{\pi }{8} = - 1 - \sqrt 2 $
We know that $\dfrac{\pi }{8}$ belongs to the first quadrant and their $\tan $ values are all positive.
Since $\sqrt 2 > 1$, we have $\sqrt 2 - 1$ is positive.
But we have $ - 1 - \sqrt 2 $ is negative.
So we get,
$\tan \dfrac{\pi }{8} = - 1 + \sqrt 2 $

Therefore the answer is option B.

Note:
Here we take the advantage that $\tan \dfrac{\pi }{4} = 1$. The important point is to identify the angle as the half of $\dfrac{\pi }{4}$ and apply the result. Thus we got a quadratic equation which could be easily solved to find the answer. Also remember that all trigonometric function has positive values on the first quadrant,