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Instead of multiplying the top and bottom of the 2nd term by cos^2 A, you can instead DIVIDE the top and bottom of the FIRST term by cos^2 A. After replacing the resulting sec^2 A by (1+tan^2 A), you get another perfect square in the numerator which cancels as before. This avoids magically...

They give you the answer in the question. When they do this, you are permitted to substitute to prove that your parametric coordinates satisfy the Cartesian equation.

sec2A+tan2A=\frac{1}{cos2A}+tan2A =\frac{1}{cos^2 A-sin^2 A} +\frac{2tanA}{1-tan^2 A}\times \frac{cos^2 A}{cos^2 A} =\frac{cos^2 A + sin^2 A}{cos^2 A-sin^2 A} +\frac{2sin A cosA}{cos^2 A-sin^2 A} =\frac{(cosA+sinA)^2}{(cosA+sinA)(cosA-sinA)} =\frac{cosA+sinA}{cosA-sinA}

I'm just wondering about how students at (fully) selective high schools judge the quality of their maths teachers, and the value of their maths teacher in relation to the value of their tutor. So I'd like to conduct a survey. Feel free to answer any or all of these questions. (1) Name your...