Question

Evaluate using trigonometric functions: ${\tan ^2}\dfrac{\pi }{{16}} + {\tan ^2}\dfrac{{2\pi }}{{16}} + {\tan ^2}\dfrac{{3\pi }}{{16}}.......{\tan ^2}\dfrac{{7\pi }}{{16}}$

Answer 1

According to the question, Rewrite the values of $\dfrac{{7\pi }}{{16}}$ , $\dfrac{{6\pi }}{{16}}$ , $\dfrac{{5\pi }}{{16}}$ in the given equation. Hence, use the trigonometric formulas and simplify to solve the equation.

Formula used:
Here we use the formula of trigonometric functions that are ${\tan ^2}\left( {\dfrac{\pi }{2} - x} \right) = {\cot ^2}x$ , $2\sin A\cos A = \sin \dfrac{A}{2}$ , ${\sin ^2}\dfrac{\pi }{{16}} + {\cos ^2}\dfrac{\pi }{{16}} = 1$ , $\left( {1 - \cos 2x} \right) = 2{\sin ^2}x$ .

Complete step-by-step answer:
Let’s start by rewriting the equation ${\tan ^2}\dfrac{\pi }{{16}} + {\tan ^2}\dfrac{{2\pi }}{{16}} + {\tan ^2}\dfrac{{3\pi }}{{16}}.......{\tan ^2}\dfrac{{7\pi }}{{16}}$ as ${\tan ^2}\dfrac{\pi }{{16}} + {\tan ^2}\dfrac{{2\pi }}{{16}} + {\tan ^2}\dfrac{{3\pi }}{{16}} + {\tan ^2}\dfrac{{4\pi }}{{16}} + {\tan ^2}\dfrac{{5\pi }}{{16}} + {\tan ^2}\dfrac{{6\pi }}{{16}} + {\tan ^2}\dfrac{{7\pi }}{{16}}$ by filling the dots with values.
For simplifying the equation we can simplify each specific part of equation. As we know that $\dfrac{{7\pi }}{{16}}$ can be written as $\dfrac{\pi }{2} - \dfrac{\pi }{{16}}$ , $\dfrac{{6\pi }}{{16}}$ can be written as $\dfrac{\pi }{2} - \dfrac{{2\pi }}{{16}}$ , $\dfrac{{5\pi }}{{16}}$ can be written as $\dfrac{\pi }{2} - \dfrac{{3\pi }}{{16}}$ and $\dfrac{{4\pi }}{{16}}$ can be written as $\dfrac{\pi }{4}$ . $\dfrac{\pi }{2} - \dfrac{{2\pi }}{{16}}$
So, on substituting all the values we get,
${\tan ^2}\dfrac{\pi }{{16}} + {\tan ^2}\dfrac{{2\pi }}{{16}} + {\tan ^2}\dfrac{{3\pi }}{{16}} + {\tan ^2}\dfrac{\pi }{4} + {\tan ^2}\left( {\dfrac{\pi }{2} - \dfrac{{3\pi }}{{16}}} \right) + {\tan ^2}\left( {\dfrac{\pi }{2} - \dfrac{{2\pi }}{{16}}} \right) + {\tan ^2}\left( {\dfrac{\pi }{2} - \dfrac{\pi }{{16}}} \right)$
And we also know that, ${\tan ^2}\left( {\dfrac{\pi }{2} - x} \right) = {\cot ^2}x$ So, using this we will replace tan by cot on all possible positions.
$\Rightarrow {\tan ^2}\dfrac{\pi }{{16}} + {\tan ^2}\dfrac{{2\pi }}{{16}} + {\tan ^2}\dfrac{{3\pi }}{{16}} + {\tan ^2}\dfrac{\pi }{4} + {\cot ^2}\left( {\dfrac{{3\pi }}{{16}}} \right) + {\cot ^2}\left( {\dfrac{{2\pi }}{{16}}} \right) + {\cot ^2}\left( {\dfrac{\pi }{{16}}} \right)$ .
Taking tan and cot with the same degree in brackets for easy solving.
$\Rightarrow \left( {{{\tan }^2}\dfrac{\pi }{{16}} + {{\cot }^2}\dfrac{\pi }{{16}}} \right) + \left( {{{\tan }^2}\dfrac{{2\pi }}{{16}} + {{\cot }^2}\dfrac{{2\pi }}{{16}}} \right) + \left( {{{\tan }^2}\dfrac{{3\pi }}{{16}} + {{\cot }^2}\dfrac{{3\pi }}{{16}}} \right) + {\tan ^2}\dfrac{\pi }{4}$ - Equation 1.
Solving the first bracket by using the formula ${a^2} + {b^2} = {\left( {a + b} \right)^2} - 2ab$ .
${\tan ^2}\dfrac{\pi }{{16}} + {\cot ^2}\dfrac{\pi }{{16}} = {\left( {\tan \dfrac{\pi }{{16}} + \cot \dfrac{\pi }{{16}}} \right)^2} - 2\tan \dfrac{\pi }{{16}}\cot \dfrac{\pi }{{16}}$ .
Converting tan and cot in terms of sin and cos. So, we get $\Rightarrow {\left( {\dfrac{{\sin \dfrac{\pi }{{16}}}}{{\cos \dfrac{\pi }{{16}}}} + \dfrac{{\cos \dfrac{\pi }{{16}}}}{{\sin \dfrac{\pi }{{16}}}}} \right)^2} - 2$ .
By Taking the LCM we get $\Rightarrow {\left( {\dfrac{{{{\sin }^2}\dfrac{\pi }{{16}} + {{\cos }^2}\dfrac{\pi }{{16}}}}{{\cos \dfrac{\pi }{{16}}\sin \dfrac{\pi }{{16}}}}} \right)^2} - 2$ .
Multiplying numerator and denominator by 2.
$\Rightarrow {\left( {\dfrac{{2*\left( {{{\sin }^2}\dfrac{\pi }{{16}} + {{\cos }^2}\dfrac{\pi }{{16}}} \right)}}{{2\cos \dfrac{\pi }{{16}}\sin \dfrac{\pi }{{16}}}}} \right)^2} - 2$.
As we know ${\sin ^2}\dfrac{\pi }{{16}} + {\cos ^2}\dfrac{\pi }{{16}}$ is equal to 1.
$\Rightarrow {\left( {\dfrac{{2*1}}{{2\cos \dfrac{\pi }{{16}}\sin \dfrac{\pi }{{16}}}}} \right)^2} - 2$
Using the identity $2\sin A\cos A = \sin \dfrac{A}{2}$
So, we get ${\left( {\dfrac{2}{{\sin \dfrac{\pi }{8}}}} \right)^2} - 2$
Opening the square,
$\Rightarrow \dfrac{4}{{{{\sin }^2}\dfrac{\pi }{8}}} - 2$
Multiplying numerator and denominator by 2.
$\Rightarrow \dfrac{{4*2}}{{2{{\sin }^2}\dfrac{\pi }{8}}} - 2$
Using the formula $\Rightarrow \left( {1 - \cos 2x} \right) = 2{\sin ^2}x$ .
$\Rightarrow \dfrac{8}{{\left( {1 - \cos \dfrac{\pi }{4}} \right)}} - 2$
Substituting value of $\cos \dfrac{\pi }{4}$ .
$\Rightarrow \dfrac{8}{{\left( {1 - \dfrac{1}{{\sqrt 2 }}} \right)}} - 2$
By Taking the LCM we get $\Rightarrow \dfrac{{8\sqrt 2 }}{{\left( {\sqrt 2 - 1} \right)}} - 2$.
Rationalising with $\left( {\sqrt 2 + 1} \right)$ , we get $\Rightarrow 8\sqrt 2 \left( {\sqrt 2 + 1} \right) - 2$.
Similarly after solving 2nd bracket $\Rightarrow \dfrac{8}{{\left( {1 - \cos \dfrac{\pi }{2}} \right)}} - 2 = 8 - 2 = 6$ and Similarly after solving 3rd bracket $\Rightarrow \dfrac{8}{{\left( {1 - \cos \dfrac{{3\pi }}{4}} \right)}} - 2 = 8\sqrt 2 \left( {\sqrt 2 - 1} \right) - 2$ .
Thus, Substituting values in equation (1) $\Rightarrow \left( {{{\tan }^2}\dfrac{\pi }{{16}} + {{\cot }^2}\dfrac{\pi }{{16}}} \right) + \left( {{{\tan }^2}\dfrac{{2\pi }}{{16}} + {{\cot }^2}\dfrac{{2\pi }}{{16}}} \right) + \left( {{{\tan }^2}\dfrac{{3\pi }}{{16}} + {{\cot }^2}\dfrac{{3\pi }}{{16}}} \right) + {\tan ^2}\dfrac{\pi }{4}$ .
$\Rightarrow 8\sqrt 2 \left( {\sqrt 2 + 1} \right) - 2 + 6 + 1 + 8\sqrt 2 \left( {\sqrt 2 - 1} \right) - 2$
$\Rightarrow 16 + 8\sqrt 2 - 2 + 7 + 8\sqrt 2 *\sqrt 2 - 8\sqrt 2 - 2$
$\Rightarrow 35$

Note: To solve these types of questions, we must remember the trigonometric formulas and algebraic identities to solve it in a simpler way. Hence, simplify all the values by taking L.C.M or rationalising to get the desired result.