Evaluate each of the following

Solution:

We have,

$\left(\operatorname{cosec}^{2} 45^{\circ} \sec ^{2} 30^{\circ}\right)\left(\sin ^{2} 30^{\circ}+4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{-}\right)$...(1)

Now,

$\sin 30^{\circ}=\frac{1}{2}, \operatorname{cosec} 45^{\circ}=\sqrt{2}, \sec 30^{\circ}=\frac{2}{\sqrt{3}}, \sec 60^{\circ}=2, \cot 45^{\circ}=1$

So by substituting above values in equation (1)

We get,

$\left(\operatorname{cosec}^{2} 45^{\circ} \sec ^{2} 30^{\circ}\right)\left(\sin ^{2} 30^{\circ}+4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}\right)$

$=\left((\sqrt{2})^{2} \times\left(\frac{2}{\sqrt{3}}\right)^{2}\right) \times\left(\left(\frac{1}{2}\right)^{2}+4 \times(1)^{2}-(2)^{2}\right)$

$=\left(2 \times \frac{(2)^{2}}{(\sqrt{3})^{2}}\right) \times\left(\frac{1^{2}}{2^{2}}+4 \times 1-4\right)$

$=\left(2 \times \frac{4}{3}\right) \times\left(\frac{1}{4}+4-4\right)$

$=\left(\frac{8}{3}\right) \times\left(\frac{1}{4}\right)$

Now, in above equation 4 cancel 8 and 2 remains

Hence,

$\left(\operatorname{cosec}^{2} 45^{\circ} \sec ^{2} 30^{\circ}\right)\left(\sin ^{2} 30^{\circ}+4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}\right)$

$=\frac{2}{3}$

Therefore,

$\left(\operatorname{cosec}^{2} 45^{\circ} \sec ^{2} 30^{\circ}\right)\left(\sin ^{2} 30^{\circ}+4 \cot ^{2} 45^{\circ}-\sec ^{2} 60^{\circ}\right)=\frac{2}{3}$