Question

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

Answer 1

$Sin(\frac{\pi }{2}+\alpha)+Cos(\pi+\alpha)+tg(\frac{3\pi }{2}-\alpha)+Ctg(2\pi -\alpha )=Cos\alpha-Cos\alpha+Ctg\alpha-Ctg\alpha=0$

Answer 2

To simplify the expression, let's use the trigonometric identities:

1. sin(π/2 + a) = cos(a)
2. cos(π + a) = -cos(a)
3. tan(3π/2 - a) = -cot(a) (cotangent is the reciprocal of tangent)
4. cot(2π - a) = -tan(a) (cotangent is the reciprocal of tangent)

Now, let's substitute these identities into the expression:

cos(a) - cos(a) - cot(a) - (-tan(a))

Now, simplify further:

cos(a) - cos(a) + tan(a) + tan(a)

Since the two pairs of terms have opposite signs, they cancel each other out, and the simplified expression is:

2tan(a)