Question

$2{\tan ^{ - 1}}\left( { - 3} \right)$ is equal to
$\left( a \right) - {\cos ^{ - 1}}\left( {\dfrac{{ - 4}}{5}} \right)$
$\left( b \right) - \pi + {\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right)$
$\left( c \right)\dfrac{{ - \pi }}{2} + {\tan ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right)$
$\left( d \right){\cot ^{ - 1}}\left( {\dfrac{4}{3}} \right)$

Answer 1

Hint: In this particular question use the concept that ${\tan ^{ - 1}}\left( { - x} \right) = - {\tan ^{ - 1}}x,$ ${\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ and also use that ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$ so use these concepts to reach the solution of the question.

Complete step by step answer:
Now we have to find out the value of:
$2{\tan ^{ - 1}}\left( { - 3} \right)$
Now as we know that ${\tan ^{ - 1}}\left( { - x} \right) = - {\tan ^{ - 1}}x,$so use this property in the above equation we have,
$ \Rightarrow 2{\tan ^{ - 1}}\left( { - 3} \right) = - 2{\tan ^{ - 1}}3$.......... (1)
Now we also know that, ${\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right),x \geqslant 0$, so use this property in the above equation we have,
So as we see that in the above equation x = 3, so we have,
$ \Rightarrow 2{\tan ^{ - 1}}\left( { - 3} \right) = - 2{\tan ^{ - 1}}3 = - {\cos ^{ - 1}}\left( {\dfrac{{1 - {3^2}}}{{1 + {3^2}}}} \right)$
Now simplify it we have,
$ \Rightarrow 2{\tan ^{ - 1}}\left( { - 3} \right) = - 2{\tan ^{ - 1}}3 = - {\cos ^{ - 1}}\left( {\dfrac{{ - 8}}{{10}}} \right) = - {\cos ^{ - 1}}\left( {\dfrac{{ - 4}}{5}} \right)$
Now as we know that ${\cos ^{ - 1}}\left( { - x} \right) = \pi - {\cos ^{ - 1}}x,x \in \left[ { - 1,1} \right]$
So in the above equation x = -4/5, therefore, $x \in \left[ { - 1,1} \right]$ so we have,
$ \Rightarrow - {\cos ^{ - 1}}\left( {\dfrac{{ - 4}}{5}} \right) = - \left( {\pi - {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right)$
$ \Rightarrow - {\cos ^{ - 1}}\left( {\dfrac{{ - 4}}{5}} \right) = - \pi + {\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right)$
Now as we know that $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$ so use this property in equation (1) we have,
$ \Rightarrow 2{\tan ^{ - 1}}\left( { - 3} \right) = - 2{\tan ^{ - 1}}3 = - {\tan ^{ - 1}}\left( {\dfrac{6}{{1 - {3^2}}}} \right) = - {\tan ^{ - 1}}\left( {\dfrac{{ - 3}}{4}} \right)$
Now as we know that ${\tan ^{ - 1}}x = {\cot ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ so apply this in the above equation we have,
$ \Rightarrow 2{\tan ^{ - 1}}\left( { - 3} \right) = - {\tan ^{ - 1}}\left( {\dfrac{{ - 3}}{4}} \right) = - {\cot ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right)$................ (2)
Now as we know that ${\cot ^{ - 1}}\left( { - x} \right) = - {\cot ^{ - 1}}x$ so we have,
$ \Rightarrow 2{\tan ^{ - 1}}\left( { - 3} \right) = - {\cot ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right) = - \left( { - {{\cot }^{ - 1}}\left( {\dfrac{4}{3}} \right)} \right) = {\cot ^{ - 1}}\left( {\dfrac{4}{3}} \right)$
Now as we know that ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$ so we have,
$ \Rightarrow {\cot ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right) = \dfrac{\pi }{2}$
$ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right) - \dfrac{\pi }{2} = - {\cot ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right)$
$ \Rightarrow - {\cot ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right) = - \dfrac{\pi }{2} + {\tan ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right)$
So from equation (2) we have,
$ \Rightarrow 2{\tan ^{ - 1}}\left( { - 3} \right) = - {\cot ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right) = - \dfrac{\pi }{2} + {\tan ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right)$

Hence options (a), (b), (c), and (d) all options are correct.

Note: Whenever we face such types of questions the key concept we have to remember is always recall all the basic inverse trigonometric identities such as, $2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right)$, ${\tan ^{ - 1}}x = {\cot ^{ - 1}}\left( {\dfrac{1}{x}} \right)$ and rest of the equations which are useful to solve this problem are stated above, so simply use them as above we will get the required answer.