Question

If we have the value of x as $\pi < x< 2\pi $ then find the value of $\dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}$.

Answer 1

Hint: We first try to use the formula of submultiple angle. Then for the root value we take modulus and based on the value of $\pi < x <2\pi $, we take their respective sign. Then using the formula of $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}$ we convert the equation into a single trigonometric function.

Complete step-by-step solution
We have the trigonometric submultiple angle formula of $1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$ and also $1-\cos x=2{{\sin }^{2}}\dfrac{x}{2}$.
We put the values in the given equation of $\dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}$ to get
$\dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}=\dfrac{\sqrt{2{{\cos }^{2}}\dfrac{x}{2}}+\sqrt{2{{\sin }^{2}}\dfrac{x}{2}}}{\sqrt{2{{\cos }^{2}}\dfrac{x}{2}}-\sqrt{2{{\sin }^{2}}\dfrac{x}{2}}}$
Now for the square root values we take their modulus values.
$\dfrac{\sqrt{2{{\cos }^{2}}\dfrac{x}{2}}+\sqrt{2{{\sin }^{2}}\dfrac{x}{2}}}{\sqrt{2{{\cos }^{2}}\dfrac{x}{2}}-\sqrt{2{{\sin }^{2}}\dfrac{x}{2}}}=\dfrac{\sqrt{2}\left| \cos \dfrac{x}{2} \right|+\sqrt{2}\left| \sin \dfrac{x}{2} \right|}{\sqrt{2}\left| \cos \dfrac{x}{2} \right|-\sqrt{2}\left| \sin \dfrac{x}{2} \right|}=\dfrac{\left| \cos \dfrac{x}{2} \right|+\left| \sin \dfrac{x}{2} \right|}{\left| \cos \dfrac{x}{2} \right|-\left| \sin \dfrac{x}{2} \right|}$
As it’s given $\pi < x< 2\pi $, multiplying $\dfrac{1}{2}$ we get $\dfrac{\pi }{2}<\dfrac{x}{2}<\pi $. In that range value of $\cos \dfrac{x}{2}< 0$ and value of $\sin \dfrac{x}{2}> 0$. The modulus values change to \[\left| \cos \dfrac{x}{2} \right|\to -\cos \dfrac{x}{2}\] and $\left| \sin \dfrac{x}{2} \right|\to \sin \dfrac{x}{2}$ in the range of $\dfrac{\pi }{2}< \dfrac{x}{2}< \pi $.
The final equation becomes $\dfrac{\left| \cos \dfrac{x}{2} \right|+\left| \sin \dfrac{x}{2} \right|}{\left| \cos \dfrac{x}{2} \right|-\left| \sin \dfrac{x}{2} \right|}=\dfrac{-\cos \dfrac{x}{2}+\sin \dfrac{x}{2}}{-\cos \dfrac{x}{2}-\sin \dfrac{x}{2}}$.
Now we take the common of $-\cos \dfrac{x}{2}$.
So, $\dfrac{-\cos \dfrac{x}{2}+\sin \dfrac{x}{2}}{-\cos \dfrac{x}{2}-\sin \dfrac{x}{2}}=\dfrac{-\cos \dfrac{x}{2}\left( 1-\tan \dfrac{x}{2} \right)}{-\cos \dfrac{x}{2}\left( 1+\tan \dfrac{x}{2} \right)}$.
We convert the value 1 into $\tan \dfrac{\pi }{4}$ and get $\dfrac{-\cos \dfrac{x}{2}\left( 1-\tan \dfrac{x}{2} \right)}{-\cos \dfrac{x}{2}\left( 1+1.\tan \dfrac{x}{2} \right)}=\dfrac{\tan \dfrac{\pi }{4}-\tan \dfrac{x}{2}}{1+\tan \dfrac{\pi }{4}.\tan \dfrac{x}{2}}$.
Then we use the formula $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A.\tan B}$
$\dfrac{\tan \dfrac{\pi }{4}-\tan \dfrac{x}{2}}{1+\tan \dfrac{\pi }{4}.\tan \dfrac{x}{2}}=\tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)$.
Therefore, the value of $\dfrac{\sqrt{1+\cos x}+\sqrt{1-\cos x}}{\sqrt{1+\cos x}-\sqrt{1-\cos x}}$ is $\tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right)$.

Note: Instead of taking $-\cos \dfrac{x}{2}$ as common we could have taken $-\sin \dfrac{x}{2}$. The function would have been of $\cot \alpha $ instead of $\tan \alpha $. For any case of root value, we always need to use the modulus value if otherwise mentioned.