If tan (A - B) = 1 and sec (A + B) = frac{2}{ | vedclass

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(B) Given $	an (A - B) = 1$. Since $	an 	heta = 1$ at $	heta = \frac{\pi}{4}$,we have $A - B = \frac{\pi}{4} \dots (i)$.Given $\sec (A + B) = \fra

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$\frac{19}{24}\pi$

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(B) Given $	an (A - B) = 1$. Since $	an 	heta = 1$ at $	heta = \frac{\pi}{4}$,we have $A - B = \frac{\pi}{4} \dots (i)$.Given $\sec (A + B) = \frac{2}{\sqrt{3}}$. Since $\sec 	heta = \frac{2}{\sqrt{3}}$ at $	heta = \frac{\pi}{6}$,we have $A + B = \frac{\pi}{6} + 2n\pi$ or $A + B = \frac{11\pi}{6} + 2n\pi$.To find the smallest positive value of $B$,we solve for $B$ using $B = \frac{(A+B) - (A-B)}{2}$.Using $A + B = \frac{11\pi}{6}$ and $A - B = \frac{\pi}{4}$:$2B = \frac{11\pi}{6} - \frac{\pi}{4} = \frac{22\pi - 3\pi}{12} = \frac{19\pi}{12}$.Thus,$B = \frac{19\pi}{24}$.

If $\tan (A - B) = 1$ and $\sec (A + B) = \frac{2}{\sqrt{3}}$,then the smallest positive value of $B$ is

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