Question

Tgx+tgy=4cosx·cosy=1/5найти:tg(x+y)

Answer 1

Ответ:

$\pm \displaystyle\frac{4}{3}$

Пошаговое объяснение:

$\[{\mathop{\rm tg}\nolimits} x + {\mathop{\rm tg}\nolimits} y = \displaystyle\frac{{\sin x}}{{\cos x}} + \displaystyle\frac{{\sin y}}{{\cos y}} = \displaystyle\frac{{\sin x\cos y + \cos x\sin y}}{{\cos x\cos y}} =$

$= \displaystyle\frac{{\sin (x + y)}}{{\cos x\cos y}} = \sin (x + y):\displaystyle\frac{1}{5} = 5\sin (x + y) = 4;$

$\sin (x + y) = \displaystyle\frac{4}{5}.$

Используя основное тригонометрическое тождество ${\sin ^2}\alpha + {\cos ^2}\alpha = 1,$ получаем

${\sin ^2}(x + y) + {\cos ^2}(x + y) = 1;$

${\cos ^2}(x + y) = 1 - {\left( {\displaystyle\frac{4}{5}} \right)^2} = 1 - \displaystyle\frac{{16}}{{25}} = \displaystyle\frac{9}{{25}};$

$\cos (x + y) = \pm \displaystyle\frac{3}{5}.$

Тогда

${\mathop{\rm tg}\nolimits} (x + y) = \displaystyle\frac{4}{5}:\left( { \pm \displaystyle\frac{3}{5}} \right) = \pm \displaystyle\frac{4}{3}.$

Answer 2

To find $\tan(x+y)$, you can use the trigonometric identity:

$\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \cdot \tan y}$

Given that $\tan x \cdot \tan y = \frac{4}{5}$, you can substitute this into the formula. Let $\tan x = t$ and $\tan y = s$ for simplicity.

$t \cdot s = \frac{4}{5}$

Now, the given equation $\tg x + \tg y = 4\cos x \cdot \cos y = \frac{1}{5}$ can be written in terms of $t$ and $s$:

$t + s = \frac{1}{5}$

Now, use these equations to find $\tan(x+y)$:

$\tan(x+y) = \frac{t + s}{1 - ts}$

Substitute the values:

$\tan(x+y) = \frac{\frac{1}{5}}{1 - \frac{4}{5}}$

Simplify:

$\tan(x+y) = \frac{\frac{1}{5}}{\frac{1}{5}}$

$\tan(x+y) = 1$

So, $\tan(x+y) = 1$.