(a) If 4 cot^2 45^°- sec^2 60^°+ sin^2 60^°+ p =3/4, then find the value of p.

\[\begin{aligned} & 4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}+\sin ^{2} 60^{\circ}+\mathrm{p}=\frac{3}{4} \\ & \Rightarrow 4(1)^{2}-(2)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}+\mathrm{p}=\frac{3}{4} \\ & \Rightarrow 4-4+\frac{3}{4}+\mathrm{p}=\frac{3}{4} \\ & \Rightarrow \mathrm{p}=0 \end{aligned} \]

\[\begin{aligned} \cos A+\cos ^{2} A=1 & \Rightarrow \cos A=1-\cos ^{2} A=\sin ^{2} A \\ \therefore \sin ^{2} A+\sin ^{4} A & =\cos A+\cos ^{2} A\left(\because \sin ^{2} A=\cos A\right) \\ & =1 \end{aligned} \]