If $\alpha=\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right)$ and $\beta=\tan ^{-1}\left(-\tan \frac{2 \pi}{3}\right)$, then
(a) $4 \alpha=3 \beta$
(b) $3 a=4 \beta$
(c) $\alpha-\beta=\frac{7 \pi}{12}$
(d) none of these
(a) $4 \alpha=3 \beta$
We know that $\tan ^{-1}(\tan x)=x$.
$\therefore \alpha=\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right)$
$=\tan ^{-1}\left\{\tan \left(\pi+\frac{\pi}{4}\right)\right\}$
$=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)$
$=\frac{\pi}{4}$
and
$\beta=\tan ^{-1}\left\{-\tan \left(\frac{2 \pi}{3}\right)\right\}$
$=\tan ^{-1}\left\{-\tan \left(\pi-\frac{\pi}{3}\right)\right\}$
$=\tan ^{-1}\left\{\tan \left(\frac{\pi}{3}\right)\right\}$
$=\frac{\pi}{3}$
$\therefore 4 \alpha=\pi$
$3 \beta=\pi$
$\therefore 4 \alpha=3 \beta$
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