Question

simplify the expression \[\tan \left[ \left( \dfrac{1}{2} \right){{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}}+\left( \dfrac{1}{2} \right){{\cos }^{-1}}\dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right]\] where \[xy\ne 1\]

Answer 1

Hint: We can solve this question by using the basic trigonometric functions and inverse of trigonometric functions formulas and properties. Here we first need to solve the sine inverse and cosine inverse separately first using double angle formulas and then solve the tangent function to get the answer needed.

Complete step-by-step solution:
To solve this expression we will first start by solving and simplifying the sine inverse function that is
\[{{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}}\]
First we let x here be equal to tangent since doing that would let us allow the double angle formula therefore let
\[x=\tan \theta \]
Therefore
\[{{\sin }^{-1}}\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }\]
Now using the identity for tangent here we know that
\[{{\sin }^{-1}}\dfrac{2\tan \theta }{{{\sec }^{2}}\theta }\]
Writing tangent and secant function in form of sine and cosine functions
\[{{\sin }^{-1}}\dfrac{2\sin \theta {{\cos }^{2}}\theta }{\cos \theta }\]
Dividing and simplifying we get
\[{{\sin }^{-1}}\sin 2\theta \]
Therefore first part is equal to \[2\theta \]
Now for the cosine function \[{{\cos }^{-1}}\dfrac{1-{{y}^{2}}}{1+{{y}^{2}}}\]
We start by letting y be tangent function too
\[{{\cos }^{-1}}\dfrac{1-{{\tan }^{2}}\alpha }{1+{{\tan }^{2}}\alpha }\]
Now using the identity for tangent here we know that
\[{{\cos }^{-1}}\dfrac{1-{{\tan }^{2}}\alpha }{{{\sec }^{2}}\alpha }\]
Writing tangent and secant function in form of sine and cosine functions
\[{{\cos }^{-1}}\dfrac{1-\dfrac{{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha }}{\dfrac{1}{{{\cos }^{2}}\alpha }}\]
Taking LCM of numerator
\[{{\cos }^{-1}}\dfrac{\dfrac{{{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha }}{\dfrac{1}{{{\cos }^{2}}\alpha }}\]
This gives us
\[{{\cos }^{-1}}\left( {{\cos }^{2}}\alpha -{{\sin }^{2}}\alpha \right)\]
Now using double angle formula we get that
\[{{\cos }^{-1}}\left( \cos 2\alpha \right)\]
Therefore the cosine part of the expression can be written as \[2\alpha \]
Now substituting these values we get
\[\tan \left[ \left( \dfrac{1}{2} \right)2\theta +\left( \dfrac{1}{2} \right)2\alpha \right]\]
\[\tan \left[ \theta +\alpha \right]\]
Now taking the sum formula of tangent we get that the expression equals to
\[\dfrac{\tan \theta +\tan \alpha }{1-\tan \theta \tan \alpha }\]
Now we know that the value of \[\theta \] and \[\alpha \] in forms of tangent are
\[x=\tan \theta \] ; \[y=\tan \alpha \]
Therefore putting the values of x and y in the expression we get that
\[\dfrac{x+y}{1-xy}\]
Hence \[\tan \left[ \left( \dfrac{1}{2} \right){{\sin }^{-1}}\dfrac{2x}{1+{{x}^{2}}}+\left( \dfrac{1}{2} \right){{\cos }^{-1}}\dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right]=\dfrac{x+y}{1-xy}\]

Note: If an equation for all values for an angle is true and the equation involves trigonometric ratios then the equation is called to be a trigonometric identity. An alternative to solve this question is that instead of substituting tan and solving it to get the double angle we can also use the sine double angle formula for tangent function.