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

Hint: 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.