Question

Tg²210°+tg²120°+tg135°=​

Answer 1

Ответ:в прикреплённом файле

Пошаговое объяснение:в прикреплённом файле

Answer 2

Let's calculate the value of the expression $\tan^2(210^\circ) + \tan^2(120^\circ) + \tan(135^\circ)$.

First, note that the tangent function has a periodicity of $180^\circ$, which means that $\tan(180^\circ + \theta) = \tan(\theta)$ for any angle $\theta$.

So, let's simplify each term:

1. $\tan^2(210^\circ)$: We can rewrite $210^\circ$ as $210^\circ = 180^\circ + 30^\circ$. Since the tangent function has a periodicity of $180^\circ$, we can write: $\tan(210^\circ) = \tan(30^\circ)$

   Now, $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, so: $\tan^2(210^\circ) = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$

2. $\tan^2(120^\circ)$: Similarly, we can rewrite $120^\circ$ as $120^\circ = 180^\circ - 60^\circ$, and since the tangent function has a periodicity of $180^\circ$, we can write: $\tan(120^\circ) = \tan(-60^\circ)$

   Now, $\tan(-60^\circ) = -\frac{\sqrt{3}}{3}$, so: $\tan^2(120^\circ) = \left(-\frac{\sqrt{3}}{3}\right)^2 = \frac{3}{3} = 1$

3. $\tan(135^\circ)$: The tangent of $135^\circ$ is equal to $-1$.

Now, let's calculate the sum:

$\tan^2(210^\circ) + \tan^2(120^\circ) + \tan(135^\circ) = \frac{1}{3} + 1 - 1 = \frac{1}{3}$

So, the value of the expression is $\frac{1}{3}$.