Question

write the function value in terms of the cofunction of a complementary angle.
tan 116°28
tan 116°28 = □(□°□)
(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 tangent is \(\tan\theta=\cot(90^{\circ}-\theta)\) when \(\theta\) is an acute angle. But for \(\theta = 116^{\circ}28'\), we know that \(\tan\theta=\tan(180^{\circ}-\alpha)=-\tan\alpha\) and also \(\tan\theta=\cot(90^{\circ}-\theta + 180^{\circ}n)\) (using periodicity and co - function relations).

First, \(\tan116^{\circ}28'=\tan(180^{\circ}- 63^{\circ}32')=-\tan63^{\circ}32'\). Also, using the co - function formula \(\tan\theta=\cot(90^{\circ}-\theta)\) in a more general form (considering the unit - circle and angle addition/subtraction).

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

\(\tan116^{\circ}28'=\cot(116^{\circ}28'- 90^{\circ})\)

Step2: Calculate the angle

Calculate \(116^{\circ}28'-90^{\circ}\)
\(116^{\circ}28'-90^{\circ}=(116 - 90)^{\circ}28'=26^{\circ}28'\)

Answer:

\(\tan116^{\circ}28'=\cot(26^{\circ}28')\)