Question

Find the value of the following: $\tan \dfrac{1}{2}\left[ {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]$, if $\left| x \right|\le 1,y\ge 0$ and $xy\le 1$.

Answer 1

Hint: Assume the value of the given expression as ‘E’ .Substitute $x=\tan \theta $ and $y=\tan \phi $. Simplify the equations: $\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ and $\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right)$ using the formulas: $\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta $ and $\left( \dfrac{1-{{\tan }^{2}}\phi }{1+{{\tan }^{2}}\phi } \right)=\cos 2\phi $. Once simplified, use the identities: ${{\sin }^{-1}}\left( \sin a \right)=a$ and ${{\cos }^{-1}}\left( \cos b \right)=b$, to get rid of inverse functions. Now, use the relation of tangent of a sum of two angles, given as: $\tan \left( \theta +\phi \right)=\dfrac{\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }$ and again substitute the value of $\tan \theta $ and $\tan \phi $ to get the answer.

Complete step-by-step solution -
We have been given, to find the value of: $\tan \dfrac{1}{2}\left[ {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]$.
Let us assume the value of this expression as ‘E’.
Substituting, $x=\tan \theta $ and $y=\tan \phi $, we get,
$E=\tan \dfrac{1}{2}\left[ {{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)+{{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\phi }{1+{{\tan }^{2}}\phi } \right) \right]$
Applying the formulas: $\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta $ and $\left( \dfrac{1-{{\tan }^{2}}\phi }{1+{{\tan }^{2}}\phi } \right)=\cos 2\phi $, we get,
$E=\tan \dfrac{1}{2}\left[ {{\sin }^{-1}}\left( \sin 2\theta \right)+{{\cos }^{-1}}\left( \cos 2\phi \right) \right]$
Now, using the identity: ${{\sin }^{-1}}\left( \sin a \right)=a$ and ${{\cos }^{-1}}\left( \cos b \right)=b$, we have,
$\begin{align}
  & E=\tan \dfrac{1}{2}\left[ 2\theta +2\phi \right] \\
 & \Rightarrow E=\tan \left[ \dfrac{2\theta +2\phi }{2} \right] \\
 & \Rightarrow E=\tan \left[ \theta +\phi \right] \\
\end{align}$
Applying the formula for tangent of a sum of two angles, given as: $\tan \left( \theta +\phi \right)=\dfrac{\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }$, we have,
$E=\dfrac{\tan \theta +\tan \phi }{1-\tan \theta \tan \phi }$
Initially we have assumed: $x=\tan \theta $ and $y=\tan \phi $, therefore, again substituting the values of $\tan \theta $ and $\tan \phi $ in the simplified expression of ‘E’, we get,
$E=\dfrac{x+y}{1-xy}$
Hence, the value of the expression, $\tan \dfrac{1}{2}\left[ {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right]$ is $\dfrac{x+y}{1-xy}$.

Note: It is important to note that we can derive the formulas used above, which are $\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }=\sin 2\theta $ and $\left( \dfrac{1-{{\tan }^{2}}\phi }{1+{{\tan }^{2}}\phi } \right)=\cos 2\phi $. Here, we have to break the tangent of the given angle into its ratio of sine and cosine and then we just have to simplify it using certain trigonometric identities. But it will be beneficial for us if we will remember these formulas. It will help in solving the problems in less time. These formulas can be directly used. Note that the conditions: $\left| x \right|\le 1,y\ge 0$ and $xy\le 1$, given in the question, represent the values of ‘x’ and ‘y’ for which the expression is defined.