The value of $$\sin^2 60^\circ \cos^2 45^\circ + 2 \tan^2 60^\circ - \cosec^2 30^\circ$$ is equal to:
$$\sin^260^{\circ}\cos^245^{\circ}+2\tan^260^{\circ}-\operatorname{cosec}^230^{\circ}=\left(\frac{\sqrt{3}}{2}\right)^2.\left(\frac{1}{\sqrt{2}}\right)^2+2\left(\sqrt{3}\right)^2-\left(2\right)^2$$
$$=\frac{3}{4}.\frac{1}{2}+2\left(3\right)-4$$
$$=\frac{3}{8}+6-4$$
$$=\frac{3}{8}+2$$
$$=\frac{3+16}{8}$$
$$=\frac{19}{8}$$
Hence, the correct answer is Option B
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