Question

If $\theta = \dfrac{\pi }{{12}}$ , then the value of ${\tan ^4}\theta - 14{\tan ^2}\theta$
A. -1
B. 0
C. 1
D. 3

Answer 1

If the value corresponding to the particular value of angle is not known, then the angle should be broken down into 2 such angles whose values are already known. Then the difference or sum of angle formula should be used. The value of $\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x.\tan y}}$ and $\tan \left( {x + y} \right) =\dfrac{{\tan x + \tan y}}{{1 - \tan x.\tan y}}$. Where x and y are those angles, whose values are known for the tangent function.

Complete step by step solution: $\dfrac{\pi }{{12}} = \dfrac{\pi }{3} - \dfrac{\pi }{4}$ ,
Where, [ $\dfrac{\pi }{{12}} = \dfrac{\pi }{{12}} \times \dfrac{{{{180}^o}}}{\pi } = {15^o}$ , $\dfrac{\pi }{3} = \dfrac{\pi }{3} \times \dfrac{{{{180}^o}}}{\pi } = {60^o}$ and $\dfrac{\pi }{4} = \dfrac{\pi }{4} \times \dfrac{{{{180}^o}}}{\pi } = {45^o}$]
To convert angle given into degree from radian multiply by$\dfrac{{{{180}^o}}}{\pi }$ .
Now we know the tangent difference identity formula as.
$\tan \left( {x - y} \right) = \dfrac{{\tan x - \tan y}}{{1 + \tan x.\tan y}}......(1)$
Here, $x = \frac{\pi }{3}$ and $y = \frac{\pi }{4}$
Substitute the value in equation (1),
$\tan \left( {\dfrac{\pi }{3} - \dfrac{\pi }{4}} \right) = \dfrac{{\tan \left( {\dfrac{\pi }{3}} \right) - \tan \left( {\dfrac{\pi }{4}} \right)}}{{1 + \tan \left( {\dfrac{\pi }{3}} \right)\tan \left( {\dfrac{\pi }{4}} \right)}}......(2)$
The value of $\tan \left( {\dfrac{\pi }{3}} \right) = \sqrt 3$ and $\tan \left( {\dfrac{\pi }{4}} \right) = 1$ , substitute it in equation (2)
\tan \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\sqrt 3 - 1}}{{1 + \left( {\sqrt 3 } \right)\left( 1 \right)}} \\\ \tan \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \\\
It should be simplified by rationalizing it, multiply numerator and denominator by $\left( {\sqrt 3 - 1} \right)$

\tan \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)}} \\\ $$ The formula for ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$ $ \tan \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2} - 2\left( {\sqrt 3 } \right)\left( 1 \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( 1 \right)}^2}}} \\\ \tan \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{3 + 1 - 2\sqrt 3 }}{{3 - 1}} \\\ \tan \left( {\dfrac{\pi }{{12}}} \right) = \dfrac{{4 - 2\sqrt 3 }}{2} \\\ \tan \left( {\dfrac{\pi }{{12}}} \right) = 2 - \sqrt 3 \\\ $ The value of ${\tan ^2}\left( {\dfrac{\pi }{{12}}} \right) = {\left( {2 - \sqrt 3 } \right)^2}$ It can be expanded by $${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$$ $$ {\left( {2 - \sqrt 3 } \right)^2} = {\left( 2 \right)^2} + {\left( {\sqrt 3 } \right)^2} - 2\left( 2 \right)\left( {\sqrt 3 } \right) \\\ {\left( {2 - \sqrt 3 } \right)^2} = 4 + 3 - 4\sqrt 3 \\\ {\left( {2 - \sqrt 3 } \right)^2} = 7 - 4\sqrt 3 \\\ $$ The expression is given as $E = {\tan ^4}\dfrac{\pi }{{12}} - 14{\tan ^2}\left( {\dfrac{\pi }{4}} \right)$ Take out ${\tan ^2}\dfrac{\pi }{4}$ common $E = {\tan ^2}\left( {\dfrac{\pi }{{12}}} \right)\left[ {{{\tan }^2}\dfrac{\pi }{{12}} - 14} \right]......(4)$ Substitute the value of ${\tan ^2}\dfrac{\pi }{{12}}$ in equation (3), $ E = \left( {7 - 4\sqrt 3 } \right)\left( {7 + 4\sqrt 3 } \right) \\\ E = \left( {7 - 4\sqrt 3 } \right)\left( {7 + 4\sqrt 3 } \right) \\\ E = - \left( {7 - 4\sqrt 3 } \right)\left( {7 + 4\sqrt 3 } \right) \\\ E = - \left( {{7^2} - {{\left( {4\sqrt 3 } \right)}^2}} \right) \\\ E = - \left( {49 - 48} \right) \\\ E = - 1 \\\ $ Hence, the correct option is (A). **Note:** The angle given in radian may be converted into degree and evaluated based on the convenience. For simplicity of calculation the angle should be broken down, based on the formula available in the form of sum and difference of tangent of angle.