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

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(B) Given $	an (A - B) = 1$. Since $	an(\pi/4) = 1$,we have $A - B = \frac{\pi}{4} + n\pi$. For the smallest positive value,consider $A - B = \frac{

Vedclass · Accepted Answer

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

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

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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