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Question
Find the value of the trigonometric functions for the following:
tan θ = −2, θ lies in the II quadrant
Solution
We know that sec2θ – tan2θ = 1
sec2θ – (– 2)2 = 1
sec2θ – 4 = 1
sec2θ = 1 + 4 = 5
sec θ = `+- sqrt(5)`
Since θ lies in the second quadrant sec θ is negative.
∴ sec θ = `- sqrt(5)`
cos θ = `1/sectheta = -1/sqrt(5)`
We know cos2θ + sin2θ = 1
`(- 1/sqrt(5))^2 + sin^2theta` = 1
`1/5 + sin^2theta` = 1
sin2θ = `1 - 1/5 = (5 - 1)/5`
sin2θ = `4/5`
sin θ = `+- 2/sqrt(5)`
Since θ lies in the second quadrant sin θ is positivee.
∴ sin θ = `2/sqrt(5)`
sin θ = `2/sqrt(5)`, cosec = `1/sintheta = sqrt(5)/2`
cos θ = `- 1/sqrt(5)`, sec θ = `1/costheta = - sqrt(5)`
tan θ = – 2, cot θ = `1/tantheta = - 1/2`
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