$x, y$ are positive acute angles with $x + y < 90^{o}$ and $sin (2 x - 20^{o}) = cos (2 y + 20^{o})$ . Then find $tan (x + y)$ .

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Solution

Given for $x, y$ are positive acute angles with $x + y < 90^{o}$ and
$sin (2 x - 20^{o}) = cos (2 y + 20^{o})$
or, $sin (2 x - 20^{o}) = sin {90^{o} - (2 y + 20^{o})}$
or, $2 x - 20^{o} = 90^{o} - (2 y + 20^{o})$
or, $2 x + 2 y = 90^{o}$
or, $x + y = 45^{o}$ .
Then
$tan (x + y)$
$= tan 45^{o}$
$= 1$ .

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