You can put this solution on YOUR website! If tan 0=2 with 0 in quadrant 111 , find sec 0?
hypotenuse of reference right triangle in quadrant III=√(2^2+1^2)=√5
cos(0)=1/√5
sec(0)=1/cos(0)=√5
You can put this solution on YOUR website! tan(t)=2 with t in quadrant 111 , find sec(t)0?
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Note: Both x and y are negative in QIII
Since tan = y/x, y = -2 and x = -1
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r = sqrt(2^2+1^2] = sqrt(5)
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Your Problem::
sec(t) = r/x = sqrt(5)/-1 = -sqrt(5)
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Cheers,
Stan H.
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