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Question: Solve the following for \(x\): \(3{\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right)...

Solve the following for xx:
3tan⁡−1(x)+cot⁡−1(x)=π3{\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi

Explanation

Solution

we are asked to solve 3tan⁡−1(x)+cot⁡−1(x)=π3{\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi . Here, we will convert the equation in one trigonometric function using identity and then we will simplify to obtain the required answer.
Formula to be used:
The required trigonometric identity that is used to solve the given problem is as follows.
tan⁡x=cot⁡(π2−x)\tan x = \cot \left( {\dfrac{\pi }{2} - x} \right)

Complete step by step answer:
The given equation is
3tan⁡−1(x)+cot⁡−1(x)=π3{\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi
⇒2tan⁡−1(x)+tan⁡−1(x)+cot⁡−1(x)=π\Rightarrow 2{\tan ^{ - 1}}\left( x \right) + {\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi
Before getting into next step, we shall consider tan⁡−1(x)=p{\tan ^{ - 1}}\left( x \right) = p
Let tan⁡−1(x)=p{\tan ^{ - 1}}\left( x \right) = p. Then, tan⁡p=x\tan p = x.
tan⁡p=cot⁡(π2−p)\tan p = \cot \left( {\dfrac{\pi }{2} - p} \right) (Here we have substituted the trigonometric identity tan⁡x=cot⁡(π2−x)\tan x = \cot \left( {\dfrac{\pi }{2} - x} \right))
⇒x=cot⁡(π2−p)\Rightarrow x = \cot \left( {\dfrac{\pi }{2} - p} \right) (We have substituted tan⁡p=x\tan p = x)
⇒cot⁡−1x=π2−p\Rightarrow {\cot ^{ - 1}}x = \dfrac{\pi }{2} - p
⇒tan⁡−1(x)+cot⁡−1(x)=p+π2−p\Rightarrow {\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = p + \dfrac{\pi }{2} - p (Here we have added both the inverse of tangent and cotangent)
⇒tan⁡−1(x)+cot⁡−1(x)=π2\Rightarrow {\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} ……..(1)\left( 1 \right)
Now, we shall get into our solution.
3tan⁡−1(x)+cot⁡−1(x)=π3{\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi
⇒2tan⁡−1(x)+tan⁡−1(x)+cot⁡−1(x)=π\Rightarrow 2{\tan ^{ - 1}}\left( x \right) + {\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \pi
⇒2tan⁡−1(x)+π2=π\Rightarrow 2{\tan ^{ - 1}}\left( x \right) + \dfrac{\pi }{2} = \pi (Here we have substituted the equation (1)\left( 1 \right))
⇒2tan⁡−1(x)=π−π2\Rightarrow 2{\tan ^{ - 1}}\left( x \right) = \pi - \dfrac{\pi }{2}
⇒2tan⁡−1(x)=π2\Rightarrow 2{\tan ^{ - 1}}\left( x \right) = \dfrac{\pi }{2}
⇒tan⁡−1(x)=π2×12\Rightarrow {\tan ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} \times \dfrac{1}{2}
⇒tan⁡−1(x)=π4\Rightarrow {\tan ^{ - 1}}\left( x \right) = \dfrac{\pi }{4}
⇒x=tan⁡π4\Rightarrow x = \tan \dfrac{\pi }{4}
⇒x=1\Rightarrow x = 1 (We know that tan⁡π4=1\tan \dfrac{\pi }{4} = 1)
Hence x=1x = 1 is the desired solution for the given equation.

Note:
Here students should not that ⇒tan⁡−1(x)+cot⁡−1(x)=π2 \Rightarrow {\tan ^{ - 1}}\left( x \right) + {\cot ^{ - 1}}\left( x \right) = \dfrac{\pi }{2} this expression is also an identity. So we can directly use it to simplify the calculation.