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Question: \(2{\tan ^{ - 1}}\left( { - 3} \right)\) is equal to \(\left( a \right) - {\cos ^{ - 1}}\left( {\d...

2tan⁡−1(−3)2{\tan ^{ - 1}}\left( { - 3} \right) is equal to
(a)−cos⁡−1(−45)\left( a \right) - {\cos ^{ - 1}}\left( {\dfrac{{ - 4}}{5}} \right)
(b)−π+cos⁡−1(45)\left( b \right) - \pi + {\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right)
(c)−π2+tan⁡−1(−43)\left( c \right)\dfrac{{ - \pi }}{2} + {\tan ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right)
(d)cot⁡−1(43)\left( d \right){\cot ^{ - 1}}\left( {\dfrac{4}{3}} \right)

Explanation

Solution

In this particular question use the concept that tan⁡−1(−x)=−tan⁡−1x,{\tan ^{ - 1}}\left( { - x} \right) = - {\tan ^{ - 1}}x, tan⁡−1x=cos⁡−1(1−x21+x2){\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right) and also use that tan⁡−1x+cot⁡−1x=π2{\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:
2tan⁡−1(−3)2{\tan ^{ - 1}}\left( { - 3} \right)
Now as we know that tan⁡−1(−x)=−tan⁡−1x,{\tan ^{ - 1}}\left( { - x} \right) = - {\tan ^{ - 1}}x,so use this property in the above equation we have,
⇒2tan⁡−1(−3)=−2tan⁡−13\Rightarrow 2{\tan ^{ - 1}}\left( { - 3} \right) = - 2{\tan ^{ - 1}}3.......... (1)
Now we also know that, tan⁡−1x=cos⁡−1(1−x21+x2),x⩾0{\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,
⇒2tan⁡−1(−3)=−2tan⁡−13=−cos⁡−1(1−321+32)\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,
⇒2tan⁡−1(−3)=−2tan⁡−13=−cos⁡−1(−810)=−cos⁡−1(−45)\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(−x)=π−cos⁡−1x,x∈[−1,1]{\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∈[−1,1]x \in \left[ { - 1,1} \right] so we have,
⇒−cos⁡−1(−45)=−(π−cos⁡−1(45))\Rightarrow - {\cos ^{ - 1}}\left( {\dfrac{{ - 4}}{5}} \right) = - \left( {\pi - {{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right)} \right)
⇒−cos⁡−1(−45)=−π+cos⁡−1(45)\Rightarrow - {\cos ^{ - 1}}\left( {\dfrac{{ - 4}}{5}} \right) = - \pi + {\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right)
Now as we know that 2tan⁡−1x=tan⁡−1(2x1−x2)2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right) so use this property in equation (1) we have,
⇒2tan⁡−1(−3)=−2tan⁡−13=−tan⁡−1(61−32)=−tan⁡−1(−34)\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⁡−1x=cot⁡−1(1x){\tan ^{ - 1}}x = {\cot ^{ - 1}}\left( {\dfrac{1}{x}} \right) so apply this in the above equation we have,
⇒2tan⁡−1(−3)=−tan⁡−1(−34)=−cot⁡−1(−43)\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(−x)=−cot⁡−1x{\cot ^{ - 1}}\left( { - x} \right) = - {\cot ^{ - 1}}x so we have,
⇒2tan⁡−1(−3)=−cot⁡−1(−43)=−(−cot⁡−1(43))=cot⁡−1(43)\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⁡−1x+cot⁡−1x=π2{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2} so we have,
⇒cot⁡−1(−43)+tan⁡−1(−43)=π2\Rightarrow {\cot ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right) + {\tan ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right) = \dfrac{\pi }{2}
⇒tan⁡−1(−43)−π2=−cot⁡−1(−43)\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right) - \dfrac{\pi }{2} = - {\cot ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right)
⇒−cot⁡−1(−43)=−π2+tan⁡−1(−43)\Rightarrow - {\cot ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right) = - \dfrac{\pi }{2} + {\tan ^{ - 1}}\left( {\dfrac{{ - 4}}{3}} \right)
So from equation (2) we have,
⇒2tan⁡−1(−3)=−cot⁡−1(−43)=−π2+tan⁡−1(−43)\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, 2tan⁡−1x=tan⁡−1(2x1−x2)2{\tan ^{ - 1}}x = {\tan ^{ - 1}}\left( {\dfrac{{2x}}{{1 - {x^2}}}} \right), tan⁡−1x=cot⁡−1(1x){\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.