Solveeit Logo
Login

Question

Question: Solve for x the given inverse trigonometric equation \({{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi ...

Solve for x the given inverse trigonometric equation tan⁡−1x+2cot⁡−1x=2π3{{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}.

Explanation

Solution

Hint: For solving this question we will use one of the important formula from inverse trigonometric functions, i.e. tan⁡−1x+cot⁡−1x=π2{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2} . After that, we will find the value of cot⁡−1x{{\cot }^{-1}}x and then, we will apply trigonometric function cot⁡\cot and use the formula cot⁡(cot⁡−1x)=x\cot \left( {{\cot }^{-1}}x \right)=x where x∈Rx \in \text{R} and result cot⁡π6=3\cot \dfrac{\pi }{6}=\sqrt{3} to get the suitable value of xx easily.

Complete step-by-step solution -
Given:
We have to find a suitable value of xx and we have the following equation:
tan⁡−1x+2cot⁡−1x=2π3{{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3}
Now, before we proceed we should know the following formulas:
tan⁡−1x+cot⁡−1x=π2 (for all x∈R).....................(1) cot⁡(cot⁡−1x)=x where x∈R....................................................(2) cot⁡π6=3............................................................(3) \begin{aligned} &{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}\text{ }\left( \text{for all }x\in R \right).....................\left( 1 \right) \\\ & \cot \left( {{\cot }^{-1}}x \right)=x\ \text{where}\ x\in \text{R} ....................................................\left( 2 \right) \\\ & \cot \dfrac{\pi }{6}=\sqrt{3}............................................................\left( 3 \right) \\\ \end{aligned}
Now, we will use the above three formulas for solving this question.
We have the following equation:
tan⁡−1x+2cot⁡−1x=2π3 ⇒tan⁡−1x+cot⁡−1x+cot⁡−1x=2π3 \begin{aligned} & {{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3} \\\ & \Rightarrow {{\tan }^{-1}}x+{{\cot }^{-1}}x+{{\cot }^{-1}}x=\dfrac{2\pi }{3} \\\ \end{aligned}
Now, we will use the formula from the equation (1) to write tan⁡−1x+cot⁡−1x=π2{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2} in the above equation. Then,
tan⁡−1x+cot⁡−1x+cot⁡−1x=2π3 ⇒π2+cot⁡−1x=2π3 ⇒cot⁡−1x=2π3−π2 ⇒cot⁡−1x=π6 \begin{aligned} & {{\tan }^{-1}}x+{{\cot }^{-1}}x+{{\cot }^{-1}}x=\dfrac{2\pi }{3} \\\ & \Rightarrow \dfrac{\pi }{2}+{{\cot }^{-1}}x=\dfrac{2\pi }{3} \\\ & \Rightarrow {{\cot }^{-1}}x=\dfrac{2\pi }{3}-\dfrac{\pi }{2} \\\ & \Rightarrow {{\cot }^{-1}}x=\dfrac{\pi }{6} \\\ \end{aligned}
Now, we will apply the trigonometric function cot⁡\cot on both sides in the above equation. Then,
cot⁡−1x=π6 ⇒cot⁡(cot⁡−1x)=cot⁡π6 \begin{aligned} & {{\cot }^{-1}}x=\dfrac{\pi }{6} \\\ & \Rightarrow \cot \left( {{\cot }^{-1}}x \right)=\cot \dfrac{\pi }{6} \\\ \end{aligned}
Now, we will use the formula from the equation (2) to write cot⁡(cot⁡−1x)=x\cot \left( {{\cot }^{-1}}x \right)=x where x∈Rx \in \text{R} and formula from the equation (3) to write cot⁡π6=3\cot \dfrac{\pi }{6}=\sqrt{3} in the above equation. Then,
cot⁡(cot⁡−1x)=cot⁡π6 ⇒x=3 \begin{aligned} & \cot \left( {{\cot }^{-1}}x \right)=\cot \dfrac{\pi }{6} \\\ & \Rightarrow x=\sqrt{3} \\\ \end{aligned}
Now, from the above result, we conclude that a suitable value of xx will be equal to 3\sqrt{3} .
Thus, if tan⁡−1x+2cot⁡−1x=2π3{{\tan }^{-1}}x+2{{\cot }^{-1}}x=\dfrac{2\pi }{3} then, the suitable value of xx will be equal to 3\sqrt{3} .

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, we should remember the formula tan⁡−1x+cot⁡−1x=π2{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2} and apply it in such questions, to solve quickly without any tough calculation. And though the question is very easy, we should avoid calculation mistakes while solving the question.