Question

How do you find sec 2x, given tan x = 5/3 and sin x< 0?

Answer 1

`-34/16`

Explanation:

Starting from tan x, you can find sec x, because of the trigonometric identity 1 + `tan^2x` = `sec^2x`
1+`(5/3)^2` =`sec^2x`
`secx`= `sqrt(34/9)`

But since x is in Quadrant II, sec x has to be negative. That's because sec x has the same sign as cos x, because sec x = 1 / cos x. We know that cos x is negative is Quadrant II, therefore so is sec x. So,`secx`= - `sqrt(34/9)` = - `sqrt(34)/3`

Since sec x and cos x are reciprocals of each other,
cos x = 1/sec x = - `3/sqrt(34)`
`cos^2x=` 9/34.......... eq (i)

Now use the identity `sin^2x+cos^2x=1` to find sin x:
`sin^2x`= `1-(9/34)` =25/34 ......eq (ii)
Again, we know that sin x is positive in Quadrant II

We know,
sec2x= `1/cos(2x)`
= `1/(cos^2x-Sin^2x)`
substituting the values,
sec2x= `1/((9/34)-(25/34)`= `-34/16`