Smallest positive values of x and y which satisfy tan (x - y) = 1,, sec (x + y) = frac{2}{{√ 3 }}

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Trigonometrical Equations

hard

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

$x = \frac{{25\pi }}{{24}},\,y = \frac{{19\pi }}{{24}}$

$x = \frac{{37\pi }}{{24}},\,y = \frac{{7\pi }}{{24}}$

$x = \frac{\pi }{4},\,y = \frac{\pi }{2}$

$a$ or $b$ both

Solution

(d) $\tan (x – y) = 1$$ \Rightarrow $ $x – y = \frac{\pi }{4},\,\frac{{5\pi }}{4}$
(Considering values which lie between $0$ and $2\pi $)
$\sec (x + y) = \frac{2}{{\sqrt 3 }}$ $ \Rightarrow $ $x + y = \frac{\pi }{6},\,\frac{{11\pi }}{6}$
(Consider values which lie between $0$ and $2\pi $)
Since $x,\,y$ are positive, therefore $x + y > x – y$
Thus we have $x + y = \frac{{11\pi }}{6}$ and $x – y = \frac{\pi }{4}$
or $x + y = \frac{{11\pi }}{6}$ and $x – y = \frac{{5\pi }}{4}$
Solving these two systems of equations, we get
$x = \frac{{25\pi }}{{24}}$ and $y = \frac{{19\pi }}{{24}}$ or $x = \frac{{37\pi }}{{24}}$ and $y = \frac{{7\pi }}{{24}}$.

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