Question

Prove the given inverse trigonometric expression ${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)=\dfrac{\pi }{4}-\dfrac{x}{2},x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$.

Answer 1

Hint: Use the half angle formula given by: $\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}$, to change the numerator. Use the algebraic identity: ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, to break the terms of numerator as a product of two terms. Now, write $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ and $1={{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}$. Using these substitutions, write, $1+\sin x={{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}}$. Cancel the common terms. Now, divide the numerator and denominator by $\cos \dfrac{x}{2}$ and use the formula: $\dfrac{1-\tan \theta }{1+\tan \theta }=\tan \left( \dfrac{\pi }{4}-\theta \right)$. Finally, use the identity: ${{\tan }^{-1}}\left( \tan \theta \right)=\theta $ where $\theta \in (\dfrac{-\pi}{2},\dfrac{\pi}{2})$, to get the answer.

Complete step-by-step solution -
We have to prove: ${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)=\dfrac{\pi }{4}-\dfrac{x}{2}$
$L.H.S={{\tan }^{-1}}\left[ \dfrac{\cos x}{1+\sin x} \right]$
Using the half angle formula given by: $\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}$, we get,
\[L.H.S={{\tan }^{-1}}\left[ \dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{1+\sin x} \right]\]
Using the algebraic identity: ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we get,
\[L.H.S={{\tan }^{-1}}\left[ \dfrac{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}{1+\sin x} \right]\]
Now, in the denominator, writing $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ and $1={{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}$, we get,
$1+\sin x={{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$. This is of the form: ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$.
Therefore, $1+\sin x={{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}}$.
This can be written as:
$1+\sin x=\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)$
Therefore, the expression becomes:
\[L.H.S={{\tan }^{-1}}\left[ \dfrac{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)} \right]\]
Cancelling the common terms, we get,
\[L.H.S={{\tan }^{-1}}\left[ \dfrac{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)} \right]\]
Dividing both numerator and denominator by $\cos \dfrac{x}{2}$, we get,
\[\begin{align}
  & L.H.S={{\tan }^{-1}}\left[ \dfrac{1-\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}}{1+\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}} \right] \\
 & ={{\tan }^{-1}}\left[ \dfrac{1-\tan \dfrac{x}{2}}{1+\tan \dfrac{x}{2}} \right] \\
\end{align}\]
Using the formula: $\dfrac{1-\tan \theta }{1+\tan \theta }=\tan \left( \dfrac{\pi }{4}-\theta \right)$, we get,
\[L.H.S={{\tan }^{-1}}\left[ \tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right) \right]\]
Now, using the identity: ${{\tan }^{-1}}\left( \tan \theta \right)=\theta $ where $\theta \in (\dfrac{-\pi}{2},\dfrac{\pi}{2})$, we get,
\[\begin{align}
  & L.H.S=\left( \dfrac{\pi }{4}-\dfrac{x}{2} \right) \\
 & =R.H.S \\
\end{align}\]

Note: One may note that it is important to convert the trigonometric functions in numerator and denominator in their half angles because we are getting a hint from R.H.S in the question that there is a term containing half of ‘x’. We can also prove the relation by taking tangents to both sides at the starting, so that the expression gets free from inverse function. Then, we have to apply the same half angle formula to simplify L.H.S. In the R.H.S side we have to simplify $\tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)$ by using the identity: $\tan \left( \dfrac{\pi }{4}-\theta \right)=\dfrac{1-\tan \theta }{1+\tan \theta }$, to get the result.