(i)[3]
Show that $\frac{dy}{dx} = \frac{1}{\sin\theta\cos^3\theta}$.
(ii)[4]
Find the equation of the tangent to the curve at the point where $\theta = \tfrac{1}{4}\pi$, giving your answer in the form $y = mx + c$.
Mathematics 9709 · AS & A Level · Differentiation
Differentiation — practice question
A curve is defined by the parametric equations $x = 1 + 2\sin^2\theta$, $y = 4\tan\theta$.
Worked solution & mark scheme
This 7-mark question has a full step-by-step worked solution and mark scheme. One marking point: “State $\dfrac{dx}{dt} = 4\sin\theta\cos\theta$, or an equivalent form” …
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