Trigonometry help - expressing as an exact value

[MathsHelpPlz]

"Given that 180 degrees < x < 270 degrees and that tan(x)=(7/24), find the exact value of sec(x)."

So I used the identity sec^2(x)=1 + tan^2(x), and substituted (7/24) for tan(x) to get:
sec(x)=Sqrt(1 + (7/24^2) )=(25/24) so I let cos(x)=(24/25), found x as 16.26020471 and proceeded to find, using the various properties such as even properties and supplimentary properties, that x=196.2602047 is the value that is in the range, which is correct. But I'm unsure how to set out my work so that I can get the exact value of 1/cos(x) as -25/24 rather than -1.041.....

Thank you for your time, I will be very grateful for any help.

*The calculators we are allowed to use wouldn't express -1.041... as a fraction.

 

[Deleted member 4993]

"Given that 180 degrees < x < 270 degrees and that tan(x)=(7/24), find the exact value of sec(x)."

So I used the identity sec^2(x)=1 + tan^2(x), and substituted (7/24) for tan(x) to get:
sec(x)= ± Sqrt(1 + (7/24^2) )= ±(25/24) so I let cos(x)=(24/25), found x as 16.26020471 and proceeded to find, using the various properties such as even properties and supplimentary properties, that x=196.2602047 is the value that is in the range, which is correct. But I'm unsure how to set out my work so that I can get the exact value of 1/cos(x) as -25/24 rather than -1.041.....

Thank you for your time, I will be very grateful for any help.

*The calculators we are allowed to use wouldn't express -1.041... as a fraction.

Why are you trying to find 'x'?

The question asked you to find the exact value of sec(x) in third quadrant.

so sec(x) = - 25/24

 

[MathsHelpPlz]

Why are you trying to find 'x'?

The question asked you to find the exact value of sec(x) in third quadrant.

so sec(x) = - 25/24

If you find x doesn't it tell you which value is within the range? How can you tell which value is in the range?

 

[MathsHelpPlz]

Why are you trying to find 'x'?

The question asked you to find the exact value of sec(x) in third quadrant.

so sec(x) = - 25/24

Oh sorry, I see, you can use the circle, thanks very much!

 

[Deleted member 4993]

If you find x doesn't it tell you which value is within the range? How can you tell which value is in the range?

Look at the unit circle.

If your "Opposite" = -24 and "Hypotenuse" = 25 - you are in the domain as stated by the problem (π < x < 3π/2)

sec(x) = H/O = -25/24.............................That's it!!!