Question
Write the value of $\tan^{-1}\Big\{\tan\Big(\frac{15\pi}{4}\Big)\Big\}.$

Answer

We have
$\tan^{-1}\Big\{\tan\Big(\frac{15\pi}{4}\Big)\Big\}$
$=\tan^{-1}\Big\{\tan\Big(4\pi-\frac{\pi}{4}\Big)\Big\}$
$=\tan^{-1}\Big\{-\tan\Big(\frac{\pi}{4}\Big)\Big\}$ $[\because\ \tan(4\pi-\text{x})=-\tan\text{x}]$
$=\tan^{-1}\Big\{\tan\Big(-\frac{\pi}{4}\Big)\Big\}$
$=-\frac{\pi}{4}$ $\big[\because\ \tan^{-1}(\tan\text{x})=\text{x}\big]$
$\therefore\ \tan^{-1}\Big\{\tan\Big(\frac{15\pi}{4}\Big)\Big\}=-\frac{\pi}{4}$

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