Question

Cos(-a)cos(180+a)/sin(-a) cos(90+a)

Answer 1

Ответ:

$\frac{ \cos(a) \cos(180 + a) }{ \sin(a) \cos(90 + a) ?}$

Answer 2

To simplify the expression $\frac{\cos(-a) \cos(180+a)}{\sin(-a) \cos(90+a)}$, we can use trigonometric identities to simplify each term:

1. $\cos(-a)$ can be simplified as $\cos(-a) = \cos(a)$ because the cosine function is an even function, which means $\cos(-x) = \cos(x)$ for all values of $x$.

2. $\cos(180+a)$ can be simplified as $\cos(180+a) = -\cos(a)$ because $\cos(180+x) = -\cos(x)$ for all values of $x$.

3. $\sin(-a)$ can be simplified as $\sin(-a) = -\sin(a)$ because the sine function is an odd function, which means $\sin(-x) = -\sin(x)$ for all values of $x$.

4. $\cos(90+a)$ can be simplified as $\cos(90+a) = \sin(a)$ because $\cos(90+x) = \sin(x)$ for all values of $x$.

Now, let's substitute these simplified expressions back into the original expression:

$$\frac{\cos(-a) \cos(180+a)}{\sin(-a) \cos(90+a)} = \frac{\cos(a) (-\cos(a))}{(-\sin(a)) \sin(a)} = \frac{-\cos^2(a)}{-\sin(a) \sin(a)} = \frac{\cos^2(a)}{\sin^2(a)}$$

So, the simplified expression is $\frac{\cos^2(a)}{\sin^2(a)}$.