If $x$ $x$ and $y$ $y$ are acute angles and $sin (2 x - 20^{\circ}) = cos (2 y + 20^{\circ})$ $sin(2x-20^@)=cos(2y+20^@)$ , what is $tan (x + y)$ $tan(x+y)$ ?

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Answer 1 · 2017-12-20T06:41:12

$tan (x + y) = 1$ $tan(x+y)=1$

$A s . sin (90^{\circ} - θ) = cos θ$ $color(blue)(Ascolor(white)(.)sin(90^@-theta)=costheta)$

$sin (2 x - 20^{\circ}) = cos (2 y + 20^{\circ})$ $sin(2x-20^@)=cos(2y+20^@)$ can be written as

or $sin (2 x - 20^{\circ}) = sin (90^{\circ} - 2 y - 20^{\circ})$ $sin(2x-20^@)=sin(90^@-2y-20^@)$

or $sin (2 x - 20^{\circ}) = sin (70^{\circ} - 2 y)$ $sin(2x-20^@)=sin(70^@-2y)$

and as $x, y and (x + y)$ $x,y" and "(x+y)$ are acute angles

or $2 x - 20^{\circ} = 70^{\circ} - 2 y$ $2x-20^@=70^@-2y$

or $2 x + 2 y = 90^{\circ}$ $2x+2y=90^@$

and so $x + y = 45^{\circ}$ $x+y=45^@$

Hence, $tan (x + y) = {tan 45}^{\circ} = {tan 45}^{\circ} = 1 x A n s .$ $tan(x+y)=tan45^@=tan45^@=1color(white)(x)color(brown)(Ans.)$

If xx and yy are acute angles and sin(2x−20∘)=cos(2y+20∘)sin(2x-20^@)=cos(2y+20^@), what is tan(x+y)tan(x+y)?