Question

If we have an expression as ${{\tan }^{-1}}x-{{\cot }^{-1}}x={{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$, $x>0$, find the value of $x$ and hence find the value of ${{\sec }^{-1}}\left( \dfrac{2}{x} \right)$.

Answer 1

We are given a trigonometric expression and we have to solve the given expression for the value of $x$. We will be using the formula ${{\cot }^{-1}}x=\dfrac{\pi }{2}-{{\tan }^{-1}}x$, to make the given expression in terms of tangent function and then we will solve for $x$. After getting the value of $x$, we will substitute the value of $x$ in ${{\sec }^{-1}}\left( \dfrac{2}{x} \right)$ and then find the required value. Hence, we will have done the required solution.

Complete step by step solution:
According to the given question, we are given a trigonometric expression in terms of tangent and cotangent function, we have to solve the expression and find the value of $x$.
The expression we have is,
${{\tan }^{-1}}x-{{\cot }^{-1}}x={{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$----(1)
To solve the expression, we will have to make the expression in terms of either of the one function only. So, we will use the formula, ${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$. We can see that the RHS of the equation (1) has the term in tangent function, so we will convert the cotangent function in the LHS to tangent function, using the above formula. We get,
$\Rightarrow {{\tan }^{-1}}x-\left( \dfrac{\pi }{2}-{{\tan }^{-1}}x \right)={{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$----(2)
Opening up the brackets in equation (2), also we know that, $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$. So, we get the new expression as,
$\Rightarrow {{\tan }^{-1}}x-\dfrac{\pi }{2}+{{\tan }^{-1}}x=\dfrac{\pi }{6}$
Adding up the similar term. We will now separate the trigonometric functions and the angles, so we will get,
$\Rightarrow 2{{\tan }^{-1}}x-\dfrac{\pi }{2}=\dfrac{\pi }{6}$
$\Rightarrow 2{{\tan }^{-1}}x=\dfrac{\pi }{6}+\dfrac{\pi }{2}$
On solving further, we get the expression as,
$\Rightarrow 2{{\tan }^{-1}}x=\dfrac{\pi +3\pi }{6}$
$\Rightarrow 2{{\tan }^{-1}}x=\dfrac{4\pi }{6}$----(3)
Reducing the RHS, we get,
$\Rightarrow 2{{\tan }^{-1}}x=\dfrac{2\pi }{3}$
$\Rightarrow {{\tan }^{-1}}x=\dfrac{\pi }{3}$----(4)
Writing the above expression in terms of $x$, we get,
$\Rightarrow x=\tan \left( \dfrac{\pi }{3} \right)$
We know that the value of $\tan \left( \dfrac{\pi }{3} \right)=\sqrt{3}$, putting this value in the above expression, we get,
$\Rightarrow x=\sqrt{3}$
So, we have the value of $x=\sqrt{3}$.
We also have to find the value of ${{\sec }^{-1}}\left( \dfrac{2}{x} \right)$,
Substituting the value of $x=\sqrt{3}$, we get,
$\Rightarrow {{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)$
$\Rightarrow \dfrac{\pi }{3}$
That is, for $\sec \left( \dfrac{\pi }{3} \right)$ we have the value as $\dfrac{2}{\sqrt{3}}$.
Therefore, the value of $x=\sqrt{3}$ and the value of ${{\sec }^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)=\dfrac{\pi }{3}$.

Note: The given expression had to be converted to either of the one function – either tangent or cotangent function in order to solve the given question. Also, the values of trigonometric functions should be known correctly, else there is a possibility of getting the answer wrong.