Question

Sin(a+6pi)ctg(5pi-a)/sin(2pi-a)ctg(5pi+a)=?

Answer 1

sin(a+6pi)ctg(5pi-a)/sin(2pi-a)ctg(5pi+a)=sina(-ctga)/{(-sina)ctga}=1

Answer 2

Let's simplify the expression step by step:

1. First, let's look at the trigonometric identities involving cotangent (ctg):

   • cot(x) = 1/tan(x)
   • cot(-x) = -cot(x)

2. Now, we can rewrite the expression using these identities:

   (sin(a + 6π) * cot(5π - a)) / (sin(2π - a) * cot(5π + a))

3. Next, let's consider the values of sine and cotangent for angles related to π and 2π:

   • sin(π) = 0
   • sin(2π) = 0
   • cot(π) = 0
   • cot(2π) = 0

4. Since sine and cotangent are both 0 for these angles, we need to determine the values of sine and cotangent for the given angles a, 5π - a, and 5π + a.

   • sin(a + 6π) = sin(a) (since adding or subtracting multiples of 2π does not change the sine value)
   • cot(5π - a) = -cot(a) (from the identity cot(-x) = -cot(x))
   • cot(5π + a) = cot(a) (since adding or subtracting multiples of 2π does not change the cotangent value)

5. Now, we can substitute these values back into the expression:

   (sin(a) * (-cot(a))) / (sin(2π - a) * cot(a))

6. Now, let's consider the values of sine and cotangent for 2π - a:

   • sin(2π - a) = sin(-a) (since sine is an odd function, sin(-x) = -sin(x))
   • cot(a) (from the previous step)

7. Substitute these values into the expression:

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

8. Now, we see that cot(a) cancels out in the numerator and denominator:

   -cot(a) / (-cot(a))

9. Finally, the two negative signs cancel out:

   1

So, the simplified expression is equal to 1:

(sin(a + 6π) * cot(5π - a)) / (sin(2π - a) * cot(5π + a)) = 1