Question

Cos² a(1+tg²a)= sin(2п+a)

Answer 1

$cos^2a\cdot (1+tg^2a)=cos^2a\cdot \frac{1}{cos^2a}=1\\\\sin(2\pi +a)=sin\, a$

Answer 2

To solve the trigonometric equation $\cos^2(a) \cdot (1 + \tan^2(a)) = \sin(2a + \pi)$, we'll first manipulate the left-hand side of the equation using trigonometric identities.

Recall the following trigonometric identities:

1. $\sin(2a) = 2 \sin(a) \cos(a)$
2. $\cos^2(a) = 1 - \sin^2(a)$
3. $\tan^2(a) = \frac{{\sin^2(a)}}{{\cos^2(a)}}$

Let's start solving the equation:

$\cos^2(a) \cdot (1 + \tan^2(a)) = \sin(2a + \pi)$

Using identity (3), we get:

$\cos^2(a) \cdot \left(1 + \frac{{\sin^2(a)}}{{\cos^2(a)}}\right) = \sin(2a + \pi)$

Simplifying:

$\cos^2(a) + \sin^2(a) = \sin(2a + \pi)$

Using identity (2), we have:

$1 = \sin(2a + \pi)$

Now, let's use the periodicity of the sine function:

$\sin(2a + \pi) = \sin(2a)$

Recall identity (1):

$\sin(2a) = 2 \sin(a) \cos(a)$

So, the equation becomes:

$1 = 2 \sin(a) \cos(a)$

Now, we can simplify this equation. Since $\sin(\pi) = 0$, we can assume that $2a + \pi = \pi$, which gives us $a = 0$. Therefore, $\sin(a) = 0$ and $\cos(a) = 1$.

Thus, the solution to the equation is $a = 0$.