Question

Cos(α-π)*tq(π+α)/sin(π/2-α)

Answer 1

$\frac{Cos( \alpha - \pi )tg( \pi + \alpha )}{Sin( \frac{ \pi }{2}- \alpha ) } = \frac{-Cos \alpha *tg \alpha }{Cos \alpha } =-tg \alpha$

Answer 2

Let's simplify the expression step by step:

1. First, we'll use some trigonometric identities:

   • cos(α-π) = -cos(α)
   • sin(π/2-α) = cos(α)

2. Substituting these values into the expression, we get:

   • (-cos(α)) * tq(π+α) / cos(α)

3. The expression tq(π+α) is not clear since "tq" is not a standard trigonometric function. It's possible that it might be a typo or a different notation. Please clarify what "tq" represents so that I can provide a complete simplification of the expression.

If "tq" is meant to be "tan," then the final simplified expression would be:

-(-cos(α)) * tan(π+α) / cos(α)

Please confirm or provide the correct notation for "tq" to proceed with the simplification.