Question

The value of $\tan \dfrac{\pi }{{16}}$
A. $\sqrt {4 + 2\sqrt 2} + (\sqrt 2 + 1)$
B. $\sqrt 4 + 2\sqrt 2 - (\sqrt 2 + 1)$
C. $\sqrt 4 - 2\sqrt 2 - (\sqrt 2 + 1)$
D. $\sqrt 4 + 2\sqrt 2 - (\sqrt 2 - 1)$

Answer 1

Here in this question you must know the value of the trigonometric function and how to use the trigonometric functions. Use the value of $\tan \theta$ and do the calculation for finding your answer. Use the identities related to $\sin \theta$ and $\cos \theta$.

Complete step-by-step answer:
We have $\tan \dfrac{\pi }{{16}}$
We can write $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Here put the values of $\theta = \dfrac{\pi }{{16}}$
$\tan \dfrac{\pi }{{16}} = \dfrac{{\sin \dfrac{\pi }{{16}}}}{{\cos \dfrac{\pi }{{16}}}}$
Now multiplying numerator and denominator by 2 and $\sin \dfrac{\pi }{{16}}$ for simplifying the function
We have
$\tan \dfrac{\pi }{{16}} = .\dfrac{2}{2}\dfrac{{\sin \dfrac{\pi }{{16}}}}{{\cos \dfrac{\pi }{{16}}}}.\dfrac{{\sin \dfrac{\pi }{{16}}}}{{\sin \dfrac{\pi }{{16}}}}$
Simplifying the function
$\tan \dfrac{\pi }{{16}} = .\dfrac{{2{{\sin }^2}\dfrac{\pi }{{16}}}}{{2\cos \dfrac{\pi }{{16}}.\sin \dfrac{\pi }{{16}}}}$
Here we know that
$\cos 2\theta = 1 - 2{\sin ^2}\theta$
$2{\sin ^2}\theta = 1 - \cos 2\theta$
and $2\sin \theta \cos \theta = \sin 2\theta$
using these two identities we can solve the function
$2{\sin ^2}\dfrac{\pi }{{16}} = 1 - \cos 2\dfrac{\pi }{{16}}$
$2\cos \dfrac{\pi }{{16}}.\sin \dfrac{\pi }{{16}} = \sin 2\dfrac{\pi }{{16}}$
now putting the values in the function
$\tan \dfrac{\pi }{{16}} = \dfrac{{1 - \cos 2\dfrac{\pi }{{16}}}}{{\sin 2\dfrac{\pi }{{16}}}}$
Dividing the $\theta$ by 2
$\tan \dfrac{\pi }{{16}} = \dfrac{{1 - \cos \dfrac{\pi }{8}}}{{\sin \dfrac{\pi }{8}}}$
We have one more identity related to $\cos \theta$and $\sin \theta$ that is
$\cos \theta = \sqrt {\dfrac{{1 + \cos 2\theta }}{2}} ,\sin \theta = \sqrt {\dfrac{{1 - \cos 2\theta }}{2}}$
Now using the identities, we get
$\tan \dfrac{\pi }{{16}} = \dfrac{{1 - \sqrt {\dfrac{{1 + \cos 2\dfrac{\pi }{8}}}{2}} }}{{\sqrt {\dfrac{{1 - \cos 2\dfrac{\pi }{8}}}{2}} }}$
Solving the above function, we get
$\tan \dfrac{\pi }{{16}} = \dfrac{{1 - \sqrt {\dfrac{{1 + \cos \dfrac{\pi }{4}}}{2}} }}{{\sqrt {\dfrac{{1 - \cos \dfrac{\pi }{4}}}{2}} }}$
Taking the L.C.M
$\tan \dfrac{\pi }{{16}} = \dfrac{{\sqrt 2 - \sqrt {1 + \cos \dfrac{\pi }{4}} }}{{\sqrt {1 - \cos \dfrac{\pi }{4}} }}$
Putting the values of $\dfrac{\pi }{4}$=$\dfrac{1}{{\sqrt 2 }}$
$\tan \dfrac{\pi }{{16}} = \dfrac{{\sqrt 2 - \sqrt {1 + \dfrac{1}{{\sqrt 2 }}} }}{{\sqrt {1 - \dfrac{1}{{\sqrt 2 }}} }}$
Taking the L.C.M and solving the equation
$\tan \dfrac{\pi }{{16}} = \dfrac{{2 - \sqrt {\sqrt 2 + 1} }}{{\sqrt {\sqrt 2 - 1} }}$
Rationalizing the above function, we get
$\tan \dfrac{\pi }{{16}} = \dfrac{{2 - \sqrt {\sqrt 2 + 1} }}{{\sqrt {\sqrt 2 - 1} }}.\dfrac{{\sqrt {\sqrt 2 + 1} }}{{\sqrt {\sqrt 2 + 1} }}$
$\tan \dfrac{\pi }{{16}} = \dfrac{{2\sqrt {\sqrt 2 + 1} - \sqrt {{{(\sqrt 2 + 1)}^2}} }}{{\sqrt {2 - 1} }}$
Open the brackets
$\tan \dfrac{\pi }{{16}} = \dfrac{{2\sqrt {\sqrt 2 + 1} - (\sqrt 2 + 1)}}{{\sqrt 1 }}$
$\tan \dfrac{\pi }{{16}} = 2\sqrt {\sqrt 2 + 1} - (\sqrt 2 + 1)$
Here we know that $\sqrt 4 = 2$, now write $\sqrt 4$on the place of $2$
$\tan \dfrac{\pi }{{16}} = \sqrt {4(\sqrt 2 + 1)} - (\sqrt 2 + 1)$
$\tan \dfrac{\pi }{{16}} = \sqrt {4\sqrt 2 + 4} - (\sqrt 2 + 1)$
Here we can write the above function
$\sqrt 4 + 2\sqrt 2 - (\sqrt 2 + 1)$
So, the value of $\tan \dfrac{\pi }{{16}} =$\sqrt 4 + 2\sqrt 2 - (\sqrt 2 + 1)$

Hence, option B is the correct option from all.

Note: Here students mostly make mistakes in the calculation part. Use the $\sin \theta$ and $\cos \theta$ function identities for solving the question. Always rationalizing the function when there are roots in the denominator. Do not get confused and make mistakes in the calculation part of roots and square. You must know the perfect square that will help you to get your correct answer. Try to make the $\theta$ like ${0^ \circ },{30^ \circ }{.45^ \circ },{60^ \circ },{90^ \circ }$. Always rationalize the equation for simplifying it.