Question

The value of $4{\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right) - {\tan ^{ - 1}}\left( {\dfrac{1}{{239}}} \right) =$
$\left( 1 \right)\pi$
$\left( 2 \right)\dfrac{\pi }{2}$
$\left( 3 \right)\dfrac{\pi }{3}$
$\left( 4 \right)\dfrac{\pi }{4}$

Answer 1

This question requires the knowledge of standard trigonometric formulae and basic algebraic identities. Here, the tangent function is given so the standard identities for tangent function are very helpful while solving the given question. Some of the important formulae are: $\left( 1 \right)$ Quotient identity: $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ . $\left( 2 \right)$ Reciprocal identity: $\tan \theta = \dfrac{1}{{\cot \theta }}$ . $\left( 3 \right)$ Pythagoras identity: $1 + {\tan ^2}\theta = {\sec ^2}\theta$ . $\left( 4 \right)$ Sum identity: $\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$ . $\left( 5 \right)$ Difference identity: $\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$ . $\left( 6 \right)$ Double angle formula: $\tan 2a = \dfrac{{2\tan a}}{{1 - {{\tan }^2}a}}$ . $\left( 7 \right)$ Half angle formulae : $\tan \left( {\dfrac{\theta }{2}} \right) = \dfrac{{1 - \cos \theta }}{{\sin \theta }}{\text{ or }}\dfrac{{\sin \theta }}{{1 + \cos \theta }}{\text{ or }} \pm \sqrt {\dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}}$ .

Complete step by step answer:
The given expression is ; $4{\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right) - {\tan ^{ - 1}}\left( {\dfrac{1}{{239}}} \right)$
It can also be written as;
= 2\left\\{ {2{{\tan }^{ - 1}}\dfrac{1}{5}} \right\\} - {\tan ^{ - 1}}\dfrac{1}{{239}}
For simplification;
$\left( {\because 2{{\tan }^{ - 1}}\left( {\dfrac{1}{5}} \right) = {{\tan }^{ - 1}}\left( {\dfrac{1}{5}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{5}} \right)} \right)$
Hence the above expression can be written as ;
$= 2\left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{5}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{5}} \right)} \right] - {\tan ^{ - 1}}\dfrac{1}{{239}}$
By the standard formula for tangent function, we know that;
$\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\dfrac{{x + y}}{{1 - xy}}} \right]$
Applying the above formula, we get;
$= 2{\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{5} + \dfrac{1}{5}}}{{1 - \left( {\dfrac{1}{5} \times \dfrac{1}{5}} \right)}}} \right) - {\tan ^{ - 1}}\dfrac{1}{{239}}$
Simplifying the above expression;
$= 2{\tan ^{ - 1}}\left( {\dfrac{{\dfrac{2}{5}}}{{1 - \dfrac{1}{{25}}}}} \right) - {\tan ^{ - 1}}\dfrac{1}{{239}}$
We know that an expression, $\because \dfrac{{\dfrac{a}{b}}}{{\dfrac{c}{d}}} = \dfrac{a}{b} \times \dfrac{d}{c}$
Using the same logic mentioned above, we get;
$= 2{\tan ^{ - 1}}\left( {\dfrac{2}{5} \times \dfrac{{25}}{{24}}} \right) - {\tan ^{ - 1}}\dfrac{1}{{239}}$
$= 2{\tan ^{ - 1}}\left( {\dfrac{5}{{12}}} \right) - {\tan ^{ - 1}}\dfrac{1}{{239}}{\text{ }}......\left( 1 \right)$
By standard trigonometric identity for tangent function, we know that;
$\Rightarrow 2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$
Expanding the first term of the equation $\left( 1 \right)$ i.e. $2{\tan ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)$ according to the above formula;
$\Rightarrow 2{\tan ^{ - 1}}\left( {\dfrac{5}{{12}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{2 \times \dfrac{5}{{12}}}}{{1 - {{\left( {\dfrac{5}{{12}}} \right)}^2}}}} \right)$
$\Rightarrow 2{\tan ^{ - 1}}\left[ {\dfrac{{\dfrac{5}{6}}}{{\dfrac{{119}}{{144}}}}} \right] = {\tan ^{ - 1}}\left( {\dfrac{5}{6} \times \dfrac{{144}}{{119}}} \right)$
$\Rightarrow 2{\tan ^{ - 1}}\left( {\dfrac{5}{{12}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{120}}{{119}}} \right)$
Now, put the value of $2{\tan ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)$ in equation $\left( 1 \right)$ ;
$= {\tan ^{ - 1}}\left( {\dfrac{{120}}{{119}}} \right) - {\tan ^{ - 1}}\left( {\dfrac{1}{{239}}} \right){\text{ }}......\left( 2 \right)$
We know that;
$= {\tan ^{ - 1}}x - {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left[ {\dfrac{{x - y}}{{1 + xy}}} \right]$
Applying the above formula to equation $\left( 2 \right)$ ;
$= {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{120}}{{119}} - \dfrac{1}{{239}}}}{{1 + \left( {\dfrac{{120}}{{119}}} \right)\left( {\dfrac{1}{{239}}} \right)}}} \right)$
Simplifying the above expression, we get ;
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{28680 - 119}}{{28441}}}}{{\dfrac{{28441 + 120}}{{28441}}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{28561}}{{28441}} \times \dfrac{{28441}}{{28561}}} \right)$
$\Rightarrow {\tan ^{ - 1}}\left( 1 \right) = {\tan ^{ - 1}}\tan \left( {\dfrac{\pi }{4}} \right)$ $\left( {\because \tan \dfrac{\pi }{4} = 1} \right)$
By the identity ;
$\because {\tan ^{ - 1}}\tan \left( x \right) = x$
Therefore, ${\tan ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{4}$
Therefore, the value of $4{\tan ^{ - 1}}\left( {\dfrac{1}{5}} \right) - {\tan ^{ - 1}}\left( {\dfrac{1}{{239}}} \right)$ is $\dfrac{\pi }{4}$ .

So, the correct answer is “Option 4”.

Note:
The values of tangent function for different angles are listed here: $\left( 1 \right)\tan {0^0} = 0$ . $\left( 2 \right)\tan \left( {\dfrac{\pi }{6}} \right){\text{ or tan}}\left( {{{30}^0}} \right) = \dfrac{1}{{\sqrt 3 }}$ . $\left( 3 \right)\tan \left( {\dfrac{\pi }{4}} \right){\text{ or }}\tan \left( {{{45}^0}} \right) = 1$ . $\left( 4 \right)\tan \left( {\dfrac{\pi }{3}} \right){\text{ or tan}}\left( {{{60}^0}} \right) = \sqrt 3$ . $\left( 5 \right)\tan \left( {\dfrac{\pi }{2}} \right){\text{ or tan}}\left( {{{90}^0}} \right) = \infty$ . Like the sine function, the tangent function is also an odd function meaning $\tan \left( { - \theta } \right) = - \tan \theta$ . An odd function is symmetric about the y-axis , i.e. $f\left( { - x} \right) = - f\left( x \right)$ . Unlike sine and tangent function, cosine is an even function i.e. $\cos \left( { - \theta } \right) = \cos \theta$ , an even function is symmetric about the x-axis i.e. $f\left( { - x} \right) = f\left( x \right)$ .