Question

Find the value of ${{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)+{{\cos }^{-1}}\left( \cos \dfrac{13\pi }{6} \right)$

Answer 1

Now first we will convert the given equation by writing $\dfrac{5\pi }{6}=\pi -\dfrac{\pi }{6}$ and $\dfrac{13\pi }{6}=2\pi +\dfrac{\pi }{6}$ . Now we know that $\tan \left( \pi -\theta \right)=-\tan \theta$ and $\cos \left( 2\pi +\theta \right)=\cos \theta$ . Hence we can convert the given expression using these results. Now in the obtained expression we can apply the property$\tan \left( -\theta \right)=-\tan \theta$ . After this we will have a simplified expression in the form of tan and cos. Now we will use the property that ${{\tan }^{-1}}\left( \tan x \right)=x;x\in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ and similarly ${{\cos }^{-1}}\left( \cos x \right)=x;x\in \left( 0,2\pi \right)$. Hence we will get the value of the given expression.

Complete step by step answer:
Now first let us consider the given expression.
We have ${{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)+{{\cos }^{-1}}\left( \cos \dfrac{13\pi }{6} \right)$
First we know that $\dfrac{5\pi }{6}\notin \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ and also $\dfrac{13\pi }{6}\notin \left( 0,2\pi \right)$ . So first we will convert the given angles to simplify the given expression.
To do so we can write $\dfrac{5\pi }{6}=\pi -\dfrac{\pi }{6}$ and $\dfrac{13\pi }{6}=2\pi +\dfrac{\pi }{6}$
Hence we get the above expression as ${{\tan }^{-1}}\left( \tan \left( \pi -\dfrac{\pi }{6} \right) \right)+{{\cos }^{-1}}\left( \cos \left( 2\pi +\dfrac{\pi }{6} \right) \right)$
Now we know the property of tan that $\tan \left( \pi -\theta \right)=-\tan \theta$ similarly we know that $\cos \left( 2\pi +\theta \right)=\cos \theta$ .
Hence using this properties we can rewrite the above expression as
${{\tan }^{-1}}\left( -\tan \left( \dfrac{\pi }{6} \right) \right)+{{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{6} \right) \right)$
Now again we know that $\tan \left( -\theta \right)=-\tan \theta$ , using this result in the above expression we get.
${{\tan }^{-1}}\left( \tan \left( -\dfrac{\pi }{6} \right) \right)+{{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{6} \right) \right)$
Now we know that $-\dfrac{\pi }{2}<-\dfrac{\pi }{6}<\dfrac{\pi }{2}$ and $0<\dfrac{\pi }{6}<2\pi$
Hence we can say $-\dfrac{\pi }{6}\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ and $\dfrac{\pi }{6}\in \left( 0,2\pi \right)$
Now we know that for all $x\in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ ${{\tan }^{-1}}\left( \tan x \right)=x$ and for all $x\in \left( 0,2\pi \right)$ ${{\cos }^{-1}}\left( \cos x \right)=x$ .
Hence using this property in the above expression we get.
$-\dfrac{\pi }{6}+\dfrac{\pi }{6}=0$

Hence finally we get the value of expression ${{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)+{{\cos }^{-1}}\left( \cos \dfrac{13\pi }{6} \right)$ is 0.

Note: Now by looking at the given expression we can make the mistake by writing
${{\cos }^{-1}}\left( \cos \dfrac{13\pi }{6} \right)=\dfrac{13\pi }{6}$ and ${{\tan }^{-1}}\left( \tan \dfrac{5\pi }{6} \right)=\dfrac{5\pi }{6}$ and hence we will get the final answer as $\dfrac{5\pi }{6}+\dfrac{13\pi }{6}=\dfrac{18\pi }{6}=3\pi$ . This will be wrong since that $\dfrac{5\pi }{6}\notin \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ and also $\dfrac{13\pi }{6}\notin \left( 0,2\pi \right)$ . Hence not that for ${{\tan }^{-1}}\left( \tan x \right)=x$ we must have $x\in \left( \dfrac{-\pi }{2},\dfrac{\pi }{2} \right)$ and similarly for ${{\cos }^{-1}}\left( \cos x \right)=x$ we must have $x\in \left( 0,2\pi \right)$ . Note that inverse function is only possible when the function is bijective, hence we have to consider the domain and codomain of the function accordingly.