Show that \(\dfrac{1+\cot^2\theta}{\cot^2\theta}=\sec^2\theta\).[1]
Write down the derivative of \(\tan\theta\) with respect to \(\theta\).[1]
Using part (a) and part (b), find the exact value of \(\displaystyle\int_0^{\frac\pi3}\left(\dfrac{1+\cot^2\theta}{\cot^2\theta}-\sin\theta\right)\,\mathrm{d}\theta\).[4]
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