Question

For any value of $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ , prove that ${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)=\dfrac{\pi }{4}-\dfrac{x}{2}$.

Answer 1

To solve this problem, we should know the transformation formulae in trigonometry. We know that $\cos x={{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)$ and $1+\sin x={{\left( \sin \left( \dfrac{x}{2} \right)+\cos \left( \dfrac{x}{2} \right) \right)}^{2}}$. Using these two formulae, we can rewrite the L.H.S by cancelling $\sin \left( \dfrac{x}{2} \right)+\cos \left( \dfrac{x}{2} \right)$ . We get the transformed L.H.S as ${{\tan }^{-1}}\left( \dfrac{\cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right)} \right)$. By cancelling $\cos \dfrac{x}{2}$, We get ${{\tan }^{-1}}\left( \dfrac{1-\tan \left( \dfrac{x}{2} \right)}{1+\tan \left( \dfrac{x}{2} \right)} \right)={{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right) \right)$. By checking the range of $\dfrac{\pi }{4}-\dfrac{x}{2}$ , we can cancel ${{\tan }^{-1}}$ and $\tan$ to get the required R.H.S.

Complete step by step answer:
We know the formula for $\cos 2y$ in terms of $\cos y$ and $\sin y$ as
$\cos 2y={{\cos }^{2}}y-{{\sin }^{2}}y\to \left( 1 \right)$.
Let us consider a function $\sin y+\cos y$ and squaring the function, we get
${{\left( \sin y+\cos y \right)}^{2}}={{\sin }^{2}}y+{{\cos }^{2}}y+2\sin y\cos y$
We know that ${{\sin }^{2}}y+{{\cos }^{2}}y=1$ and $2\sin y\cos y=\sin 2y$.
Using them, we get
${{\left( \sin y+\cos y \right)}^{2}}=1+\sin 2y\to \left( 2 \right)$
Substituting $y=\dfrac{x}{2}$ in equations-1 and 2, we get
$\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}\to \left( 3 \right)$
$1+\sin x={{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}\to \left( 4 \right)$
Using the equations-3, 4 in ${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)$, we get
${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)={{\tan }^{-1}}\left( \dfrac{{{\cos }^{2}}\left( \dfrac{x}{2} \right)-{{\sin }^{2}}\left( \dfrac{x}{2} \right)}{{{\left( \cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right) \right)}^{2}}} \right)$
We know the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$,
Using that in above equation, we get
${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\left( \cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right) \right)\times \left( \cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right) \right)}{{{\left( \cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right) \right)}^{2}}} \right)$
We can cancel $\left( \cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right) \right)$ in numerator and denominator as $\left( \cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right) \right)\ne 0$ in the given range of x. We get
${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\cos \left( \dfrac{x}{2} \right)-\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right)} \right)$
Dividing by $\cos \left( \dfrac{x}{2} \right)$ in numerator and denominator, we get
${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)={{\tan }^{-1}}\left( \dfrac{1-\dfrac{\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)}}{1+\dfrac{\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)}} \right)$
We know that $\dfrac{\sin x}{\cos x}=\tan x$, we can write the above equation as,
${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)={{\tan }^{-1}}\left( \dfrac{1-\tan \left( \dfrac{x}{2} \right)}{1+\tan \left( \dfrac{x}{2} \right)} \right)$
We know that $\tan \dfrac{\pi }{4}=1$. We can rewrite the above equation as
${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\tan \dfrac{\pi }{4}-\tan \left( \dfrac{x}{2} \right)}{1+\tan \dfrac{\pi }{4}\times \tan \left( \dfrac{x}{2} \right)} \right)$
We know that $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1-\tan A\times \tan B}$, we can write the above equation By taking $A=\dfrac{\pi }{4},B=\dfrac{x}{2}$ as
${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)={{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right) \right)\to \left( 5 \right)$
We can write ${{\tan }^{-1}}\left( \tan x \right)=x$ when $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$.
Let us consider $\dfrac{\pi }{4}-\dfrac{x}{2}$,
When $x=-\dfrac{\pi }{2}$, we get $\dfrac{\pi }{4}-\dfrac{x}{2}=\dfrac{\pi }{4}+\dfrac{\pi }{4}=\dfrac{\pi }{2}$
When $x=\dfrac{\pi }{2}$, we get $\dfrac{\pi }{4}-\dfrac{x}{2}=\dfrac{\pi }{4}-\dfrac{\pi }{4}=0$.
For the given range of x in the question which is $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$, we can infer that
$\left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)\in \left( 0,\dfrac{\pi }{2} \right)$
As $\left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)\in \left( 0,\dfrac{\pi }{2} \right)$, we can write
${{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right) \right)=\dfrac{\pi }{4}-\dfrac{x}{2}$.
We can write equation-5 as
${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)=\dfrac{\pi }{4}-\dfrac{x}{2}$
$\therefore$For the given range of $x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$, proved that ${{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)=\dfrac{\pi }{4}-\dfrac{x}{2}$.

Note:
Students can make a mistake while using the relation ${{\tan }^{-1}}\left( \tan x \right)=x$. In this question, the values of x are within the range where the formula is valid, but we have to be careful while applying it and we have to check whether x is in the range of $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ or not. If the values of x are outside the range, we should write the corresponding value of angle within the range of $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ which has a value of tan function equal to $\tan x$. Students should check if a term is zero or not in the given range while cancelling it in numerator and denominator, like we did for $\left( \cos \left( \dfrac{x}{2} \right)+\sin \left( \dfrac{x}{2} \right) \right)$.