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# trig identities help

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Algebraically determine the exact value of sec(11pi/6+cot(8pi/3)/sin(-150). Simplify your expression completely.

$$\frac{sec\frac{11\pi}{6}+cot\frac{8\pi}{3}}{sin(-150^\circ)}$$

Jun 5, 2021
edited by Melody  Jun 5, 2021

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It is very strange to have half the question in radians and half in degrees...

$$\frac{sec\frac{11\pi}{6}+cot\frac{8\pi}{3}}{sin(-150^\circ)}\\ \frac{sec\frac{-\pi}{6}+cot\frac{2\pi}{3}}{sin(360-150^\circ)}\\$$

I really should not be doing this.

I think you either understand none of it in which case my answer will not help.

OR you are being lazy and I am just doing your homework for you.

Show us what you have done for yourself. And what exactly is troubling you.

Jun 5, 2021