Question

(ctg(-п/6)cos(-п/6)sinп/4)^-1​

Answer 1

$(Ctg(-\frac{\pi }{6})*Cos(-\frac{\pi }{6})*Sin\frac{\pi }{4})^{-1}=(-Ctg\frac{\pi }{6}*Cos\frac{\pi }{6}*Sin\frac{\pi }{4})^{-1}=(-\sqrt{3}*\frac{\sqrt{3}}{2}*\frac{\sqrt{2} }{2})^{-1}=(-\frac{3\sqrt{2} }{4})^{-1}=-\frac{4}{3\sqrt{2}}=-\frac{4*\sqrt{2}}{3\sqrt{2}*\sqrt{2}}=-\frac{4\sqrt{2}}{6}=-\frac{2\sqrt{2}}{3}$

Answer 2

To simplify the expression, we'll first evaluate the individual trigonometric functions and then proceed with the inverse. Let's start step-by-step:

1. Evaluate trigonometric functions:

   • cotangent of -π/6: cot(-π/6) = 1/tan(-π/6) = 1/(-√3) = -√3/3
   • cosine of -π/6: cos(-π/6) = cos(π/6) since cosine is an even function and remains unchanged.
   • sine of π/4: sin(π/4) = 1/√2

2. Combine the values: (cot(-π/6) * cos(-π/6) * sin(π/4))^-1 = (-√3/3 * cos(π/6) * 1/√2)^-1 = (-√3/3 * √3/2 * 1/√2)^-1 = (-√3/6)^-1

3. Take the inverse: The reciprocal of a number is simply 1 divided by that number. So, (-√3/6)^-1 = 1 / (-√3/6)

Now, to rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator, which is (√3/6).

1 / (-√3/6) * (√3/6) / (√3/6) = -√3 / 3

Therefore, the simplified expression is -√3/3.