Respuesta :

calculista
we know that

sec²θ=tan²θ+1-------> tan²θ=> sec²θ-1
 sec θ = (√37)/6

tan²θ=> [(√37)/6]²-1--------> (37/36)-1-------> 1/36
tanθ=√(1/36)--------> 1/6

remember that
 sec θ = 1/ cos θ 
the cos θ is > 0 
and
the sin θ < 0
so the angle θ belong to the fourth quadrant-------> tan θ is negative

therefore

the answer is
tan θ=-1/6
kdreams2020

Answer:

tanx = [tex]-\frac{1}{6}[/tex]

Step-by-step explanation:

hey there,

< You have to memorize trigonometric identities for this. Here's the one we'll use for this problem.

[tex]1 + tan^2x=sec^2x[/tex]

Plug in the sec since we already know what it is.

[tex]tan^2x = (\frac{\sqrt{37} }{6})^2 -1[/tex]

When you square it, you get 37/36. 37/36-1 = 1/36

tan^2x = 1/36

Since it's squared, let's find the square root of 1/36.

It would be +- 1/6.

Since sinx < 0, it has to be -1/6. >

Hope this helped! Feel free to ask anything else.

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