How do you find the values of the six trigonometric functions given secθ=−2 and sinθ>0?
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since,secx=-2
therefore,cosx=1/secx=-1/2. ...(1)
or cos^2x=1-sin^2x=1/4
i.e; sin^2x=(1-1/4)
=>sin^2x=3/4
=>sinx=√3/4...(not -√3/4... since sinx>0
sinx=√3/4. ....(2)
therefore,cosecx=4/√3. ...(3)
tanx=sinx/cosx
=(√3/4)/(-1/2)
=-√3/2. ...(4)
therefore cotx=-2/√3.
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