Question

write the function value in terms of the cofunction of a complementary angle.
cot 141°39
cot 141°39 = □(□°□)
(simplify your answers. type the cofunction abbreviation in the first input box. type the degree part of the angle measure in the second input box. type the minute part of the angle
measure in the third input box.)

Explanation:

Step1: Use co - function identity

The co - function identity for cotangent is \(\cot\theta=\tan(90^{\circ}-\theta)\) when \(\theta\) is an acute angle. But for \(\theta = 141^{\circ}39'\), we first note that \(\cot\theta=\cot(180^{\circ}-\alpha)=-\cot\alpha\) where \(\alpha = 38^{\circ}21'\) (\(180^{\circ}-141^{\circ}39'=38^{\circ}21'\)), and also \(\cot\theta=\tan(90^{\circ}-\theta + 180^{\circ}n)\) (using the general co - function formula \(\cot x=\tan(90^{\circ}-x + 180^{\circ}n)\), \(n\in\mathbb{Z}\)).

Another way: \(\cot\theta=\tan(90^{\circ}-\theta)\) can be extended using the periodicity of trigonometric functions (\(\cot x=\cot(x + 180^{\circ}n)\)).

We know that \(\cot141^{\circ}39'=\cot(90^{\circ}+ 51^{\circ}39')\). Using the co - function formula \(\cot(A + 90^{\circ})=-\tan A\), but if we use the formula \(\cot\theta=\tan(90^{\circ}-\theta+180^{\circ})\) (since the co - function formula \(\cot x=\tan(90^{\circ}-x)\) holds for acute \(x\), and using the period of \(\tan x\) which is \(180^{\circ}\)).

\(\cot141^{\circ}39'=\tan(90^{\circ}-141^{\circ}39'+180^{\circ})\)

First, calculate \(90^{\circ}-141^{\circ}39'=- 51^{\circ}39'\), then \(-51^{\circ}39'+180^{\circ}=128^{\circ}21'\)

Alternatively, using the formula \(\cot\theta=-\cot(180^{\circ}-\theta)\) and then co - function.

We know that \(\cot\theta=\tan(90^{\circ}-\theta + 180^{\circ})\)

\(90^{\circ}-141^{\circ}39'=-51^{\circ}39'\), \(-51^{\circ}39'+180^{\circ}=128^{\circ}21'\)

Another approach: \(\cot141^{\circ}39'=\cot(90^{\circ}+51^{\circ}39')\), and \(\cot(A + 90^{\circ})=-\tan A\), but if we consider the co - function formula with the period of \(\tan\) (period \(180^{\circ}\))

We use the formula \(\cot x=\tan(90^{\circ}-x+180^{\circ})\)

\(x = 141^{\circ}39'\), \(90^{\circ}-141^{\circ}39'=-51^{\circ}39'\), \(-51^{\circ}39'+180^{\circ}=128^{\circ}21'\)

We can also use the identity \(\cot\theta=\tan(270^{\circ}-\theta)\) (since \(\cot\theta=\frac{\cos\theta}{\sin\theta}\) and \(\tan(270^{\circ}-\theta)=\frac{\sin(270^{\circ}-\theta)}{\cos(270^{\circ}-\theta)}=\frac{-\cos\theta}{-\sin\theta}=\cot\theta\))

\(270^{\circ}-141^{\circ}39' = 128^{\circ}21'\)

Answer:

\(\cot141^{\circ}39'=\tan(128^{\circ}21')\)