The smallest positive values of x and y which | vedclass

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(D) Given $	an(x - y) = 1$,we have $x - y = \frac{\pi}{4}, \frac{5\pi}{4}, \dots$ (considering positive values).Given $\sec(x + y) = \frac{2}{\sqrt{3

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$a$ or $b$ both

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(D) Given $	an(x - y) = 1$,we have $x - y = \frac{\pi}{4}, \frac{5\pi}{4}, \dots$ (considering positive values).Given $\sec(x + y) = \frac{2}{\sqrt{3}}$,we have $\cos(x + y) = \frac{\sqrt{3}}{2}$,so $x + y = \frac{\pi}{6}, \frac{11\pi}{6}, \dots$ (considering positive values).Since $x, y > 0$,we require $x + y > x - y$.Case $1$: $x + y = \frac{11\pi}{6}$ and $x - y = \frac{\pi}{4}$. Adding gives $2x = \frac{22\pi + 3\pi}{12} = \frac{25\pi}{12} \Rightarrow x = \frac{25\pi}{24}$. Then $y = \frac{11\pi}{6} - \frac{25\pi}{24} = \frac{44\pi - 25\pi}{24} = \frac{19\pi}{24}$.Case $2$: $x + y = \frac{11\pi}{6}$ and $x - y = \frac{5\pi}{4}$. Adding gives $2x = \frac{22\pi + 15\pi}{12} = \frac{37\pi}{12} \Rightarrow x = \frac{37\pi}{24}$. Then $y = \frac{11\pi}{6} - \frac{37\pi}{24} = \frac{44\pi - 37\pi}{24} = \frac{7\pi}{24}$.Both pairs are valid positive solutions.

The smallest positive values of $x$ and $y$ which satisfy $\tan (x - y) = 1$ and $\sec (x + y) = \frac{2}{\sqrt{3}}$ are

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