ધારો કે tan ^{-1}left(tan frac{5 pi}{4}right) | vedclass

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(B) આપેલ છે,$\alpha = 	an ^{-1}\left(	an \frac{5 \pi}{4}ight)$.$\frac{5 \pi}{4} = \pi + \frac{\pi}{4}$ હોવાથી,$	an \frac{5 \pi}{4} = 	an \frac{\

Vedclass · Accepted Answer

$4 \alpha - 3 \beta = 0$

Vedclass · Answer

(B) આપેલ છે,$\alpha = 	an ^{-1}\left(	an \frac{5 \pi}{4}ight)$.$\frac{5 \pi}{4} = \pi + \frac{\pi}{4}$ હોવાથી,$	an \frac{5 \pi}{4} = 	an \frac{\pi}{4}$ થાય.તેથી,$\alpha = 	an ^{-1}\left(	an \frac{\pi}{4}ight) = \frac{\pi}{4}$.આથી $4 \alpha = \pi$  $(1)$.આપેલ છે,$\beta = 	an ^{-1}\left(-	an \frac{2 \pi}{3}ight)$.$	an \frac{2 \pi}{3} = 	an \left(\pi - \frac{\pi}{3}ight) = -	an \frac{\pi}{3}$ હોવાથી,$-	an \frac{2 \pi}{3} = 	an \frac{\pi}{3}$ થાય.તેથી,$\beta = 	an ^{-1}\left(	an \frac{\pi}{3}ight) = \frac{\pi}{3}$.આથી $3 \beta = \pi$  $(2)$.સમીકરણ $(1)$ અને $(2)$ પરથી,$4 \alpha = \pi$ અને $3 \beta = \pi$ મળે છે.તેથી,$4 \alpha = 3 \beta$,જેનો અર્થ છે કે $4 \alpha - 3 \beta = 0$.

ધારો કે $\tan ^{-1}\left(\tan \frac{5 \pi}{4}\right) = \alpha$ અને $\tan ^{-1}\left(-\tan \frac{2 \pi}{3}\right) = \beta$. તો:

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