91Ó°ÊÓ

Cofunction identities are a fundamental concept in trigonometry. These identities show the relationship between pairs of trigonometric functions evaluated at complementary angles.
Complementary angles add up to \ \ \( \displaystyle \frac{\pi}{2} \ \ \) radians or 90 degrees.
For instance:

• \( \sin(\theta) = \cos\left(\frac{\pi}{2} - \theta\right) \)
• \( \tan(\theta) = \cot\left(\frac{\pi}{2} - \theta\right) \)
• \( \sec(\theta) = \csc\left(\frac{\pi}{2} - \theta\right) \)

In this exercise, we specifically use the cofunction identity for tangent and cotangent:
\( \cot(\theta) = \tan\left(\frac{\pi}{2} - \theta\right) \)
This identity is key to solving the problem where we express \( \cot \frac{9 \pi}{10} \) in terms of a tangent function.

Complementary angles are two angles whose measures add up to \( \displaystyle \frac{\pi }{2} \) radians or 90 degrees. In the context of trigonometric functions, complementary angles are frequently used to apply cofunction identities. To find the complementary angle of \(\displaystyle \frac{9\pi}{10}\):
We subtract it from \( \displaystyle \frac{\pi}{2} \).

For example, \(\frac{\pi}{2} - \frac{9 \pi}{10}\):
First, convert \( \frac{\pi}{2} \) to a common denominator: \( \frac{5\pi}{10} \).
Now perform the subtraction:
\( \frac{5 \pi}{10} - \frac{9 \pi}{10} = -\frac{4 \pi}{10} = -\frac{2 \pi}{5} \).
Hence, the complementary angle of \( \frac{9 \pi}{10} \) is \( -\frac{2 \pi}{5} \).

In mathematics, a function \( f(x) \) is considered odd if it satisfies: \( f(-x) = -f(x) \).
Odd functions have symmetric properties with respect to the origin.
Among trigonometric functions, sine and tangent are classic examples of odd functions:

• \( \sin(-\theta) = -\sin(\theta) \)
• \( \tan(-\theta) = -\tan(\theta) \)

In our problem, we use the property of the tangent function being odd:
\( \tan \left( -\frac{2 \pi}{5} \right) = -\tan \left( \frac{2 \pi}{5} \right) \).
This helps us simplify the expression obtained using the cofunction identity.
Eventually, we find our answer:
\( \cot \frac{9 \pi}{10} = -\tan \left( \frac{2 \pi}{5} \right) \)