Question

The value of \[{\rm{anti}}{\log _5}\left[ {\dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}}} \right]\] is equal to
A) Odd number
B) Even numbers
C) Prime number
D) Composite number

Answer 1

Hint:
Here, we will first simplify the terms in the bracket by using the trigonometric co ratios. Then we will substitute the values of the angle and simplify the expression further. Then we will further simplify the equation by using basic mathematical operation. We will then use the antilogarithmic formula to find the value of the antilogarithmic expression.

Formula Used:
We will use the following formula:
1) Trigonometric Co-ratio: \[\tan x = \dfrac{1}{{\cot x}}\]
2) Trigonometric Ratio \[\tan \dfrac{\pi }{5} = \sqrt {5 - 2\sqrt 5 } \];\[\tan \dfrac{{2\pi }}{5} = \dfrac{5}{{\sqrt {5 - 2\sqrt 5 } }}\]
3) Antilogarithm Formula \[{\rm{anti lo}}{{\rm{g}}_b}a = {b^a}\]

Complete step by step solution:
We are given an anti logarithmic expression \[{\rm{anti}}{\log _5}\left[ {\dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}}} \right]\].
Now, let us consider only the terms inside the bracket \[\dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}}\].
We know that Trigonometric Co-ratio: \[\tan x = \dfrac{1}{{\cot x}}\]
By using the trigonometric co-ratio in the above equation, we get
\[ \Rightarrow \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}} = \dfrac{{{{\left( {\tan \left( {\dfrac{\pi }{5}} \right)} \right)}^2} + {{\left( {\tan \left( {\dfrac{{2\pi }}{5}} \right)} \right)}^2} + 20}}{{{{\left( {\dfrac{1}{{\tan \left( {\dfrac{\pi }{5}} \right)}}} \right)}^2} + {{\left( {\dfrac{1}{{\tan \left( {\dfrac{{2\pi }}{5}} \right)}}} \right)}^2} + 28}}\]
We know that the trigonometric ratio \[\tan \dfrac{\pi }{5} = \sqrt {5 - 2\sqrt 5 } \] and \[\tan \dfrac{{2\pi }}{5} = \dfrac{5}{{\sqrt {5 - 2\sqrt 5 } }}\].
Therefore, by using the trigonometric ratio in the above equation, we get

\[ \Rightarrow \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}} = \dfrac{{{{\left( {\sqrt {5 - 2\sqrt 5 } } \right)}^2} + {{\left( {\sqrt {\dfrac{5}{{\left( {5 - 2\sqrt 5 } \right)}}} } \right)}^2} + 20}}{{{{\left( {\dfrac{1}{{\sqrt {5 - 2\sqrt 5 } }}} \right)}^2} + {{\left( {\dfrac{1}{{\sqrt {\dfrac{5}{{\left( {5 - 2\sqrt 5 } \right)}}} }}} \right)}^2} + 28}}\]
Simplifying the equation, we get
\[ \Rightarrow \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}} = \dfrac{{\left( {5 - 2\sqrt 5 } \right) + \left( {\dfrac{5}{{5 - 2\sqrt 5 }}} \right) + 20}}{{\left( {\dfrac{1}{{5 - 2\sqrt 5 }}} \right) + \left( {\dfrac{{5 - 2\sqrt 5 }}{5}} \right) + 28}}\]
Now, by taking L.C.M to both the numerators and the denominators, we get
\[ \Rightarrow \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}} = \dfrac{{\dfrac{{\left( {5 - 2\sqrt 5 } \right)\left( {5 - 2\sqrt 5 } \right) + 5 + 20\left( {5 - 2\sqrt 5 } \right)}}{{\left( {5 - 2\sqrt 5 } \right)}}}}{{\dfrac{{5 + \left( {5 - 2\sqrt 5 } \right)\left( {5 - 2\sqrt 5 } \right) + 28 \times 5\left( {5 - 2\sqrt 5 } \right)}}{{5\left( {5 - 2\sqrt 5 } \right)}}}}\]
Now, by cancelling out the like terms and rewriting the equation, we get
\[ \Rightarrow \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}} = \dfrac{{5\left[ {\left( {5 - 2\sqrt 5 } \right)\left( {5 - 2\sqrt 5 } \right) + 5 + 20\left( {5 - 2\sqrt 5 } \right)} \right]}}{{5 + \left( {5 - 2\sqrt 5 } \right)\left( {5 - 2\sqrt 5 } \right) + 28 \times 5\left( {5 - 2\sqrt 5 } \right)}}\]
Now, by multiplying the binomials using the FOIL method, we get
\[ \Rightarrow \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}} = \dfrac{{5\left[ {25 - 10\sqrt 5 - 10\sqrt 5 + 20 + 5 + 100 - 40\sqrt 5 } \right]}}{{5 + 25 - 10\sqrt 5 - 10\sqrt 5 + 20 + 140\left( {5 - 2\sqrt 5 } \right)}}\]

Now, by rewriting the equation, we get
\[ \Rightarrow \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}} = \dfrac{{5\left[ {150 - 60\sqrt 5 } \right]}}{{750 - 300\sqrt 5 }}\]
Now, by rewriting the equation, we get
\[ \Rightarrow \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}} = \dfrac{{750 - 300\sqrt 5 }}{{750 - 300\sqrt 5 }}\]
Now, by cancelling out the numerator and the denominator, we get
\[ \Rightarrow \dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}} = 1\] …………………………………………………………………………..\[\left( 1 \right)\]
Now, by substituting the equation \[\left( 1 \right)\] in antilogarithmic expression, we get
\[{\rm{anti}}{\log _5}\left[ {\dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}}} \right] = {\rm{anti lo}}{{\rm{g}}_5}\left( 1 \right){\rm{ }}\].
We know that antilogarithm formula\[{\rm{anti lo}}{{\rm{g}}_b}a = {b^a}\]
Now, by using the antilogarithm formula, we get
\[ \Rightarrow {\rm{anti}}{\log _5}\left[ {\dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}}} \right] = {5^1}\].
\[ \Rightarrow {\rm{anti}}{\log _5}\left[ {\dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}}} \right] = 5\].
Therefore, the value of \[{\rm{anti}}{\log _5}\left[ {\dfrac{{{{\tan }^2}\left( {\dfrac{\pi }{5}} \right) + {{\tan }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 20}}{{{{\cot }^2}\left( {\dfrac{\pi }{5}} \right) + {{\cot }^2}\left( {\dfrac{{2\pi }}{5}} \right) + 28}}} \right]\] is\[5\].
Therefore 5 is an odd number and a prime number.

Thus, options (A) and (C) are the correct answer.

Note:
We know that Trigonometric Equation is defined as an equation involving the trigonometric ratios. Trigonometric identity is an equation which is always true for all the variables. We should know that we have many trigonometric identities which are related to all the other trigonometric equations. Trigonometric Ratios of a Particular angle are the ratios of the sides of a right angled triangle with respect to any of its acute angle. A logarithm is defined as the power to which number which must be raised to get some values. We should also know that antilogarithm is the inverse of the logarithm transform.