Evaluate :

Question.

Evaluate :

(i) $\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$

(ii) $2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$

(iii) $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$

(iv) $\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$

(v) $\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$

Solution:

(i) $\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$

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

$=\frac{3}{4}+\frac{1}{4}=1$

(ii) $2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$

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

$=2+\frac{3}{4}-\frac{3}{4}=2$

(iii) $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$

$=\frac{1 / \sqrt{2}}{\frac{2}{\sqrt{3}}+2}=\frac{1 / \sqrt{2}}{2\left(\frac{1+\sqrt{3}}{\sqrt{3}}\right)}=\frac{1(\sqrt{3})}{2 \sqrt{2}(1+\sqrt{3})}$

$=\frac{\sqrt{3}}{2(\sqrt{2})} \times \frac{(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{\sqrt{3}(\sqrt{3}-1)}{2 \sqrt{2} \times 2}$

$=\frac{(3-\sqrt{3})}{4 \sqrt{2}}$

(iv) $\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosecti} 0^{\circ}}{\sec 30^{\circ}-\cos 60^{\circ}+\cot 45^{\circ}}$

$=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}=\frac{\frac{\sqrt{3}+2 \sqrt{3}-4}{2 \sqrt{3}}}{\frac{4+\sqrt{3}+2 \sqrt{3}}{2 \sqrt{3}}}$

$=\frac{(3 \sqrt{3}-4)}{(4+3 \sqrt{3})} \frac{(4-3 \sqrt{3})}{(4-3 \sqrt{3})}$

$=\frac{12 \sqrt{3}-27-16+12 \sqrt{3}}{16-9 \times 3}=\frac{24 \sqrt{3}-43}{-11}=\frac{43-24 \sqrt{3}}{11}$

(v) $\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}$

$=\frac{5\left(\cos 60^{\circ}\right)^{2}+4\left(\sec 30^{\circ}\right)^{2}-\left(\tan 45^{\circ}\right)^{2}}{\left(\sin 30^{2}\right)^{2}+\left(\cos 30^{2}\right)^{2}}$'

$=\frac{5\left(\frac{1}{2}\right)^{2}+4\left(\frac{2}{\sqrt{3}}\right)^{2}-(1)^{2}}{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\frac{\frac{5}{4}+4 \times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}$

$=\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{5}{4}+\frac{16}{3}-1$

$=\frac{15+64-12}{12}=\frac{67}{12}$