Question

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

Answer 1

Hint: To solve this problem, we should know the transformation formulae of inverse trigonometric functions. We know that the formulae related to the inverse trigonometric formulae are
$\begin{align}
  & \sin \left( \dfrac{2x}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x\text{ if }\left| x \right|\le 1 \\
 & {{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right)=-2{{\tan }^{-1}}y\text{ if }y<0 \\
\end{align}$
Using these relations, we can get the expression in the question in the form of
$\tan \left( {{\tan }^{-1}}x-{{\tan }^{-1}}y \right)$. We know the formula ${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$. Using this formula we can get the required answer.

Complete step-by-step solution:
Let us consider the term ${{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}}$.
We can write the transformed formula for ${{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}}$ as
$\sin \left( \dfrac{2x}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x$.
We can infer that the interval in which the formula is valid is given by $\left| x \right|\le 1$ which coincides with the interval of x given in the question.
Let us consider the term ${{\cos }^{-1}}\dfrac{1-{{y}^{2}}}{1+{{y}^{2}}}$.
We can write the transformed formula for ${{\cos }^{-1}}\dfrac{1-{{y}^{2}}}{1+{{y}^{2}}}$ as
${{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right)=-2{{\tan }^{-1}}y$
We can infer that the interval in which the formula is valid is given by $y<0$ which coincides with the interval of y given in the question.
So, we can write the expression in the question as
$\tan \left( \dfrac{1}{2}\left[ {{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}}+{{\cos }^{-1}}\dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right] \right)=\tan \left( \dfrac{1}{2}\left[ 2{{\tan }^{-1}}x-2{{\tan }^{-1}}y \right] \right)$
Cancelling two in the expression, we get
$\tan \left( \dfrac{1}{2}\left[ 2{{\tan }^{-1}}x-2{{\tan }^{-1}}y \right] \right)=\tan \left( {{\tan }^{-1}}x-{{\tan }^{-1}}y \right)$
We know the formula
${{\tan }^{-1}}x-{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)$
Using this formula, we get

$\tan \left( {{\tan }^{-1}}x-{{\tan }^{-1}}y \right)=\tan \left( {{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right) \right)$
We know that $\tan \left( {{\tan }^{-1}}x \right)=x\text{ }\forall x\in R$.
We can write the above equation as
$\tan \left( \dfrac{1}{2}\left[ {{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}}+{{\cos }^{-1}}\dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right] \right)=\dfrac{x-y}{1+xy}$



Note:Students make a mistake in applying the formula of ${{\cos }^{-1}}\dfrac{1-{{y}^{2}}}{1+{{y}^{2}}}$. We should be aware of the range of values of y for which the function is defined. If y > 0, we can write the formula as
${{\cos }^{-1}}\dfrac{1-{{y}^{2}}}{1+{{y}^{2}}}=2{{\tan }^{-1}}y$.
But we are asked in the range of y < 0, in this range the formula changes as
${{\cos }^{-1}}\dfrac{1-{{y}^{2}}}{1+{{y}^{2}}}=-2{{\tan }^{-1}}y$
Students should be careful about the range of the variables while dealing with inverse trigonometric functions as they vary with the range in which they are defined in the question.