Let $x, y$ are acute with $x + y < 90^{o}$ and $sin (2 x - 20^{o}) = cos (2 y + 20^{o})$ then find the value of $tan (x + y)$ .

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Solution

Given,
$sin (2 x - 20^{o}) = cos (2 y + 20^{o})$
or, $sin (2 x - 20^{o}) = sin {90^{o} - (2 y + 20^{o})}$ [ Since $x, y$ are acute]
or, $(2 x - 20^{o}) = 90^{o} - 2 y - 20^{o}$
or, $2 x + 2 y = 90^{o}$
or, $x + y = 45^{o}$ .
Now, $tan (x + y)$
$= tan 45^{o}$
$= 1$ .

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Let x,y are acute with x+y<90o and sin(2x−20o)=cos(2y+20o) then find the value of tan(x+y).

Given,sin(2x−20o)=cos(2y+20o)or, sin(2x−20o)=sin{90o−(2y+20o)} [ Since x,y are acute]or, (2x−20o)=90o−2y−20oor, 2x+2y=90oor, x+y=45o.Now, tan(x+y)=tan45o=1.

Let $x, y$ are acute with $x + y < 90^{o}$ and $sin (2 x - 20^{o}) = cos (2 y + 20^{o})$ then find the value of $tan (x + y)$ .

Given,
$sin (2 x - 20^{o}) = cos (2 y + 20^{o})$
or, $sin (2 x - 20^{o}) = sin {90^{o} - (2 y + 20^{o})}$ [ Since $x, y$ are acute]
or, $(2 x - 20^{o}) = 90^{o} - 2 y - 20^{o}$
or, $2 x + 2 y = 90^{o}$
or, $x + y = 45^{o}$ .
Now, $tan (x + y)$
$= tan 45^{o}$
$= 1$ .