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OpenStudy (anonymous):
3tan^2 y- sec y-1=0
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OpenStudy (tkhunny):
Do you know a relationship between tangent and secant?
OpenStudy (akashdeepdeb):
Use the relationship: \( 1 + tan^2 \theta = sec^2 \theta\) You'l get a quadratic equation. Solve it. Understood? :)
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