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प्रश्न
Find the value of the trigonometric functions for the following:
cos θ = `- 1/2`, θ lies in the III quadrant
उत्तर
We know sin2θ + cos2θ = 1
`sin^2theta + (-1/2)^2` = 1
`sin^2theta + 1/4` = 1
sin2θ = `1 - 1/4 = 3/4`
sin θ = `+- sqrt(3)/2`
sin θ = `- sqrt(3)/2`,
cosec θ = `- 2/sqrt(3)`
tan θ = `sintheta/costheta = (-sqrt(3)/2)/(- 1/2) = sqrt(3)`
cot θ = `1/tantheta = 1/sqrt(3)`
sec θ = `1/costheta = 1/(-1/2)` = – 2
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