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If $4 \cot^2 45^\circ - \sec^2 60^\circ + \sin^2 60^\circ + p = \frac{3}{4}$, then find the value of p.
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$4 \cot^2 45^\circ - \sec^2 60^\circ + \sin^2 60^\circ + p = \frac{3}{4}$
$\Rightarrow 4(1)^2 - (2)^2 + (\frac{\sqrt{3}}{2})^2 + p = \frac{3}{4}$
$\Rightarrow 4 - 4 + \frac{3}{4} + p = \frac{3}{4}$
$\Rightarrow p = 0$
$\Rightarrow 4(1)^2 - (2)^2 + (\frac{\sqrt{3}}{2})^2 + p = \frac{3}{4}$
$\Rightarrow 4 - 4 + \frac{3}{4} + p = \frac{3}{4}$
$\Rightarrow p = 0$