Question

Sin(π+a)cos(3π/2-a)tg(a-π/2)/cos(π/2+a)cos(3π/2+a)tg(π+a)

Answer 1

I hope this helps you

Answer 2

To simplify the expression sin(π+a)cos(3π/2-a)tg(a-π/2)/cos(π/2+a)cos(3π/2+a)tg(π+a), let's break it down step by step:

1. We'll start by simplifying each trigonometric term individually:

   a. sin(π+a) = -sin(a) (because sin(π + θ) = -sin(θ))

   b. cos(3π/2-a) = sin(a) (because cos(3π/2 - θ) = sin(θ))

   c. tg(a-π/2) = -cot(a) (because tg(θ - π/2) = -cot(θ))

   d. cos(π/2+a) = sin(a) (because cos(π/2 + θ) = sin(θ))

   e. cos(3π/2+a) = -cos(a) (because cos(3π/2 + θ) = -cos(θ))

   f. tg(π+a) = -tan(a) (because tg(π + θ) = -tan(θ))

2. Now, we can substitute these simplifications back into the original expression:

   (-sin(a) * sin(a) * (-cot(a))) / (sin(a) * (-cos(a)) * (-tan(a)))

3. We can simplify this further:

   (-sin(a) * sin(a) * (-cot(a))) / (sin(a) * cos(a) * tan(a))

4. Now, cancel out common factors:

   -1 / (cos(a))

So, the simplified expression is -1/cos(a).