Question

use the trigonometric function values of the quadrantal angles to evaluate. 6 cos 270°+2 sec 0°+2(sec 180°)^2
6 cos 270°+2 sec 0°+2(sec 180°)^2=
(simplify your answer. type an integer or a fraction.)

Explanation:

Step1: Recall cosine value of 270°

We know that $\cos270^{\circ}=0$.

Step2: Recall secant value of 0°

Since $\sec\theta=\frac{1}{\cos\theta}$ and $\cos0^{\circ} = 1$, then $\sec0^{\circ}=1$.

Step3: Recall secant value of 180°

Since $\cos180^{\circ}=- 1$, then $\sec180^{\circ}=\frac{1}{\cos180^{\circ}}=-1$.

Step4: Substitute values into the expression

Substitute the values into $6\cos270^{\circ}+2\sec0^{\circ}+2(\sec180^{\circ})^2$.
We get $6\times0 + 2\times1+2\times(-1)^2$.

Step5: Calculate the result

$6\times0=0$, $2\times1 = 2$, $2\times(-1)^2=2\times1 = 2$.
Then $0 + 2+2=4$.

Answer:

$4$