Let tan ^{-1}left(tan frac{5 pi}{4}right) = a | vedclass

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(B) Given,$\alpha = 	an ^{-1}\left(	an \frac{5 \pi}{4}ight)$.Since $\frac{5 \pi}{4} = \pi + \frac{\pi}{4}$,we have $	an \frac{5 \pi}{4} = 	an \f

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$4 \alpha - 3 \beta = 0$

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(B) Given,$\alpha = 	an ^{-1}\left(	an \frac{5 \pi}{4}ight)$.Since $\frac{5 \pi}{4} = \pi + \frac{\pi}{4}$,we have $	an \frac{5 \pi}{4} = 	an \frac{\pi}{4}$.Thus,$\alpha = 	an ^{-1}\left(	an \frac{\pi}{4}ight) = \frac{\pi}{4}$.This implies $4 \alpha = \pi$  $(1)$.Given,$\beta = 	an ^{-1}\left(-	an \frac{2 \pi}{3}ight)$.Since $	an \frac{2 \pi}{3} = 	an \left(\pi - \frac{\pi}{3}ight) = -	an \frac{\pi}{3}$,we have $-	an \frac{2 \pi}{3} = 	an \frac{\pi}{3}$.Thus,$\beta = 	an ^{-1}\left(	an \frac{\pi}{3}ight) = \frac{\pi}{3}$.This implies $3 \beta = \pi$  $(2)$.From equations $(1)$ and $(2)$,we have $4 \alpha = \pi$ and $3 \beta = \pi$.Therefore,$4 \alpha = 3 \beta$,which means $4 \alpha - 3 \beta = 0$.

Let $\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right) = \alpha$ and $\tan ^{-1}\left(-\tan \frac{2 \pi}{3}\right) = \beta$. Then:

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