Question

Pls help !!!!!!!!!!!!

Answer 1

`C.\quad (√(17))/(6)`

Explanation:

We are given

`\tan\theta = - (√(19))/(√(17)) \\\\`

Square both sides

`\tan^2\theta = \left(- (√(19))/(√(17)) \right)^2 = (19)/(17)`

`\tan \theta = (\sin\theta)/(\cos\theta)\\\\\tan^2 \theta = (\sin^2\theta)/(\cos^2\theta)\\`

Substituting for
`tan^2 \theta` and switching sides we get

`(\sin^2\theta)/(\cos^2\theta) = (19)/(17)\\\\\text{Add 1 to both sides:}\\\\ \\ (\sin^2\theta)/(\cos^2\theta) + 1 = (19)/(17) + 1 \\\`

`(\sin^2\theta + cos^2\theta)/(\cos^2\theta) = (19 + 17)/(17) = (36)/(17)`

`\sin^2\theta + cos^2\theta = 1`

giving

`(1)/(\cos^2\theta) = (36)/(17)`

`(17)/(36) = \cos^2\theta\\\\ \cos^2\theta = (17)/(36)\\\\ \cos\theta = \sqrt{(17)/(36)}\\\\ \cos\theta = (√(17))/(6) \\\\`

This is choice C. You had it right