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Question
Find the value of the trigonometric functions for the following:
sec θ = `13/5`, θ lies in the IV quadrant
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Solution
We know that sec2θ – tan2θ = 1
`(13/2)^2 - tan^2theta` = 1
`169/25 - 1` = tan2θ
tan2θ = `(169 - 25)/25 = 144/25`
tan θ = `+- 12/5`
Since θ lies in the fourth quadrant tan θ is negative.
∴ tan θ = `- 12/5`
cos θ = `1/sectheta = 5/13`
We know sin2θ + cos2θ = 1
`sin^2theta + (5/13)^2` = 1
sin2θ = `1 - 25/169`
sin2θ = `(169 - 25)/169 = 144/169`
sin θ = `+- 12/13`
Since θ lies in the fourth quadrant sin θ is negative.
∴ sin θ = `- 12/13`
sin θ = `- 12/13`, cosec θ = `1/sintheta = - 13/12`
cos θ = `5/13`, sec θ = `1/costheta = 13/5`
tan θ = `- 12/5`, cot θ = `1/tantheta = - 5/12`
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