Question

The value of $ 2{{\tan }^{-1}}\left( \operatorname{cosec}{{\tan }^{-1}}x-\tan {{\cot }^{-1}}x \right) $ is
A. $ {{\tan }^{-1}}x $
B. $ \tan x $
C. $ \cot x $
D. $ cose{{c}^{-1}}x $

Answer 1

Hint: To solve this question we have to substitute the value of $ {{\tan }^{-1}}x=\theta $ , so we get $ x=\tan \theta $ . Then solving further we get the value of $ \theta $ in terms of x. Then, by substituting the value in the given expression and solving further we get the desired answer. Trigonometric properties are used in these questions to substitute the values.

Complete step by step answer:
We have been given an expression $ 2{{\tan }^{-1}}\left( \operatorname{cosec}{{\tan }^{-1}}x-\tan {{\cot }^{-1}}x \right) $
We have to find the value of the given expression.
Let us assume $ {{\tan }^{-1}}x=\theta $ , so we get $ x=\tan \theta $
Now, we know that $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ and $ \cos \theta =\sqrt{1-{{\sin }^{2}}\theta } $
Now, substituting the values we get
\[\begin{align}
  & \Rightarrow x=\dfrac{\sin \theta }{\cos \theta } \\
 & \Rightarrow x=\dfrac{\sin \theta }{\sqrt{1-{{\sin }^{2}}\theta }} \\
 & \Rightarrow x=\dfrac{1}{cosec\theta \sqrt{1-\dfrac{1}{cose{{c}^{2}}\theta }}} \\
 & \Rightarrow x=\dfrac{1}{cosec\theta \sqrt{\dfrac{cose{{c}^{2}}\theta -1}{cose{{c}^{2}}\theta }}} \\
 & \Rightarrow x=\dfrac{1}{\sqrt{cose{{c}^{2}}\theta -1}} \\
 & \Rightarrow \theta =cose{{c}^{-1}}\dfrac{\sqrt{1+{{x}^{2}}}}{x} \\
\end{align}\]
Now, substituting the values in given expression we get
 $ \Rightarrow 2{{\tan }^{-1}}\left[ \operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}\dfrac{\sqrt{1+{{x}^{2}}}}{x} \right)-\tan \left( {{\tan }^{-1}}\dfrac{1}{x} \right) \right] $
Now, solving further we get
 $ \begin{align}
  & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}}{x}-\dfrac{1}{x} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{x}^{2}}}-1}{x} \right) \\
\end{align} $
Now, again putting the value $ x=\tan \theta $ we get
 $ \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } \right) $
Now, we know that $ 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $
Now, substituting the values we get
 $ \begin{align}
  & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{\sqrt{{{\sec }^{2}}\theta }-1}{\tan \theta } \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{\sec \theta -1}{\tan \theta } \right) \\
\end{align} $
Now, we know that $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ and $ \sec \theta =\dfrac{1}{\cos \theta } $ .
Substituting the values we get
 $ \begin{align}
  & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{\dfrac{1}{\cos \theta }-1}{\dfrac{\sin \theta }{\cos \theta }} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{\dfrac{1}{\cos \theta }-1}{\dfrac{\sin \theta }{\cos \theta }} \right) \\
\end{align} $
Now, solving further we get
 $ \begin{align}
  & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{\dfrac{1-\cos \theta }{\cos \theta }}{\dfrac{\sin \theta }{\cos \theta }} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{2{{\sin }^{2}}\dfrac{\theta }{2}}{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\left( \dfrac{\sin \dfrac{\theta }{2}}{\cos \dfrac{\theta }{2}} \right) \\
 & \Rightarrow 2{{\tan }^{-1}}\left( \tan \dfrac{\theta }{2} \right) \\
\end{align} $
Now, we get
\[\begin{align}
  & \Rightarrow 2\times \dfrac{\theta }{2} \\
 & \Rightarrow \theta \\
\end{align}\]
Now, we have $ {{\tan }^{-1}}x=\theta $ .
So, the value of $ 2{{\tan }^{-1}}\left( \operatorname{cosec}{{\tan }^{-1}}x-\tan {{\cot }^{-1}}x \right) $ is $ {{\tan }^{-1}}x $ .
Option A is the correct answer.

Note:
 To solve these types of questions students must have a good knowledge of trigonometric ratios, trigonometric identities, and trigonometric functions. Please be careful while substituting the values. Avoid calculation mistakes. If students try to solve this question without substituting the value $ {{\tan }^{-1}}x=\theta $, they will not be able to solve it and it becomes lengthy.