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Question: The value of \(2{\cot ^{ - 1}}\dfrac{1}{2} - {\cot ^{ - 1}}\dfrac{4}{3}\) is- A.\( - \dfrac{\pi }{...

The value of 2cot⁡−112−cot⁡−1432{\cot ^{ - 1}}\dfrac{1}{2} - {\cot ^{ - 1}}\dfrac{4}{3} is-
A.−π8 - \dfrac{\pi }{8}
B.3π2\dfrac{{3\pi }}{2}
C.π4\dfrac{\pi }{4}
D.π2\dfrac{\pi }{2}

Explanation

Solution

First we will use the trigonometric identitycot⁡−1x=tan⁡−11x{\cot ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{1}{x} to convert the trigonometric function cot θ\theta to tanθ\theta . Then we know that when x>1x > 1 then 2tan⁡−1x=π+tan⁡−12x1−x22{\tan ^{ - 1}}x = \pi + {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}} and in the function, then angle is greater than one. So put x=22in the formula and put the value in the given equation. Then we will use tan⁡(−θ)=−tan⁡θ\tan \left( { - \theta } \right) = - \tan \theta to simplify the equation. We know thattan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\dfrac{{a + b}}{{1 - ab}} so we will use this formula to simplify the given equation further.

Complete step-by-step answer:
We have to find value of 2cot⁡−112−cot⁡−1432{\cot ^{ - 1}}\dfrac{1}{2} - {\cot ^{ - 1}}\dfrac{4}{3}
We know that cot⁡−1x=tan⁡−11x{\cot ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{1}{x}
Then we can write,
⇒2tan⁡−111/2−tan⁡−114/3\Rightarrow 2{\tan ^{ - 1}}\dfrac{1}{{1/2}} - {\tan ^{ - 1}}\dfrac{1}{{4/3}}
On simplifying, we get-
⇒2tan⁡−12−cot⁡−143\Rightarrow 2{\tan ^{ - 1}}2 - {\cot ^{ - 1}}\dfrac{4}{3}
We know that if x>1x > 1 then 2tan⁡−1x=π+tan⁡−12x1−x22{\tan ^{ - 1}}x = \pi + {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}
Here the given function is 2tan⁡−122{\tan ^{ - 1}}2 where x=22 which is greater than one.
On applying this formula we get-
⇒π+tan⁡−12×21−(2×2)−tan⁡−134\Rightarrow \pi + {\tan ^{ - 1}}\dfrac{{2 \times 2}}{{1 - \left( {2 \times 2} \right)}} - {\tan ^{ - 1}}\dfrac{3}{4}
On simplifying, we get-
⇒π+tan⁡−141−4−tan⁡−134\Rightarrow \pi + {\tan ^{ - 1}}\dfrac{4}{{1 - 4}} - {\tan ^{ - 1}}\dfrac{3}{4}
On further simplifying, we get-
⇒π+tan⁡−14−3−tan⁡−134\Rightarrow \pi + {\tan ^{ - 1}}\dfrac{4}{{ - 3}} - {\tan ^{ - 1}}\dfrac{3}{4}
Now we know thattan⁡(−θ)=−tan⁡θ\tan \left( { - \theta } \right) = - \tan \theta so we can write-
⇒π+(−tan⁡−143)−tan⁡−134\Rightarrow \pi + \left( { - {{\tan }^{ - 1}}\dfrac{4}{3}} \right) - {\tan ^{ - 1}}\dfrac{3}{4}
On simplifying, we get-
⇒π−tan⁡−143−tan⁡−134\Rightarrow \pi - {\tan ^{ - 1}}\dfrac{4}{3} - {\tan ^{ - 1}}\dfrac{3}{4}
Now, on taking the negative sign common we get-
⇒π−(tan⁡−143+tan⁡−134)\Rightarrow \pi - \left( {{{\tan }^{ - 1}}\dfrac{4}{3} + {{\tan }^{ - 1}}\dfrac{3}{4}} \right)-- (i)
Now we know thattan⁡−1a+tan⁡−1b=tan⁡−1a+b1−ab{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\dfrac{{a + b}}{{1 - ab}}
So on applying this formula, we get-
⇒π−(tan⁡−143+341−43.34)\Rightarrow \pi - \left( {{{\tan }^{ - 1}}\dfrac{{\dfrac{4}{3} + \dfrac{3}{4}}}{{1 - \dfrac{4}{3}.\dfrac{3}{4}}}} \right)
On simplifying, we get-
⇒π−(tan⁡−116+9121−1)\Rightarrow \pi - \left( {{{\tan }^{ - 1}}\dfrac{{\dfrac{{16 + 9}}{{12}}}}{{1 - 1}}} \right)
This will give us zero number in the denominator and we know that10\dfrac{1}{0} is not defined so we write it as 10=∞\dfrac{1}{0} = \infty
Then we get-
⇒π−tan⁡−1∞\Rightarrow \pi - {\tan ^{ - 1}}\infty
Now we know that tan⁡π2=∞\tan \dfrac{\pi }{2} = \infty
Then we can write-
⇒π−tan⁡−1(tan⁡π2)\Rightarrow \pi - {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{2}} \right)
We know that tan⁡−1(tan⁡θ)=θ{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta
Then, we can write-
⇒π−π2\Rightarrow \pi - \dfrac{\pi }{2}
On taking LCM, we get-
⇒2π−π2=π2\Rightarrow \dfrac{{2\pi - \pi }}{2} = \dfrac{\pi }{2}
The correct answer is option D.

Note: Here, we can also solve the question this way-
After eq. (i), we can write-
⇒π−(cot⁡−134+tan⁡−134)\Rightarrow \pi - \left( {{{\cot }^{ - 1}}\dfrac{3}{4} + {{\tan }^{ - 1}}\dfrac{3}{4}} \right)
Now we know thatcot⁡−1x+tan⁡−1x=π2{\cot ^{ - 1}}x + {\tan ^{ - 1}}x = \dfrac{\pi }{2} so we can write-
⇒(cot⁡−134+tan⁡−134)=π2\Rightarrow \left( {{{\cot }^{ - 1}}\dfrac{3}{4} + {{\tan }^{ - 1}}\dfrac{3}{4}} \right) = \dfrac{\pi }{2}
Then on putting this value in the equation above, we get-
⇒π−π2\Rightarrow \pi - \dfrac{\pi }{2}
On solving, we get-
⇒2π−π2=π2\Rightarrow \dfrac{{2\pi - \pi }}{2} = \dfrac{\pi }{2}
Hence option D is the correct answer.