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Evaluate the following:
(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) \( \frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}} \).
To do:
We have to evaluate the given expressions.
Solution:
(i) We know that,
$\sin 60^{\circ}=\frac{\sqrt3}{2}$
$\sin 30^{\circ}=\frac{1}{2}$
$\cos 60^{\circ}=\frac{1}{2}$
$\cos 30^{\circ}=\frac{\sqrt3}{2}$
Therefore,$ \sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}=\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}+\frac{1}{2}\times\frac{1}{2}$
$=\frac{3}{4}+\frac{1}{4}$
$=\frac{3+1}{4}$
$=\frac{4}{4}$
$=1$
Hence, $ \sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}=1$.
(ii) We know that,
$\tan 45^{\circ}=1$
$\sin 60^{\circ}=\frac{\sqrt3}{2}$
$\cos 30^{\circ}=\frac{\sqrt3}{2}$
Therefore,$ 2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}=2
(1)^2+(\frac{\sqrt3}{2})^2-(\frac{\sqrt3}{2})^2$
$=2+\frac{3}{4}-\frac{3}{4}$
$=2$
Hence, $2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}=2$.
(iii) We know that,
$\cos 45^{\circ}=\frac{1}{\sqrt2}$
$\sec 30^{\circ}=\frac{2}{\sqrt3}$
$\operatorname{cosec} 30^{\circ}=2$
Therefore,$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}=\frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}}+\frac{2}{1}}$
$=\frac{\frac{1}{\sqrt{2}}}{\frac{2+2 \sqrt{3}}{\sqrt{3}}}$
$=\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1+\sqrt{3})}$
Multiplying numerator and denominator by $(1-\sqrt{3})$ we get,
$=\frac{\sqrt{3}}{2 \sqrt{2}(1+\sqrt{3})} \times \frac{(1-\sqrt{3})}{(1-\sqrt{3})}$
$=\frac{\sqrt{3}(1-\sqrt{3})}{2 \sqrt{2}(1-3)}$
$=\frac{\sqrt{3}(1-\sqrt{3})}{2 \sqrt{2}(-2)}$
$=\frac{\sqrt{3}(1-\sqrt{3})}{-4 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{-\sqrt{6}(1-\sqrt{3})}{8}$
$=\frac{-\sqrt{6}+\sqrt{18}}{8}$
$=\frac{-\sqrt{6}+3 \sqrt{2}}{8}$
(iv) We know that,
$\sin 30^{\circ}=\frac{1}{2}$
$\tan 45^{\circ}=1$
$\operatorname{cosec} 60^{\circ}=\frac{2}{\sqrt3}$
$\sec 30^{\circ}=\frac{2}{\sqrt3}$
$\cos 60^{\circ}=\frac{1}{2}$
$cot 45^{\circ}=1$
Therefore,$\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}=\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}$
$=\frac{\frac{\sqrt{3}+2 \sqrt{3}-4}{2 \sqrt{3}}}{\frac{4+\sqrt{3}+2 \sqrt{3}}{2 \sqrt{3}}}$
$=\frac{3 \sqrt{3}-4}{2 \sqrt{3}} \times \frac{2 \sqrt{3}}{3 \sqrt{3}+4}$
On multiplying the numerator and denominator by $(3 \sqrt{3}-4)$, we get,
$=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4} \times \frac{(3 \sqrt{3}-4)}{(3 \sqrt{3}-4)}$
$=\frac{27+16-24 \sqrt{3}}{27-16}$
$=\frac{43-24 \sqrt{3}}{11}$.
(v) We know that,
$\cos 60^{\circ}=\frac{1}{2}$
$\sec 30^{\circ}=\frac{2}{\sqrt3}$
$tan 45^{\circ}=1$
$\sin 30^{\circ}=\frac{1}{2}$
$\cos 30^{\circ}=\frac{\sqrt3}{2}$
Therefore,$\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}= \frac{5(\frac{1}{2})^{2}+4(\frac{2}{\sqrt{3}})^{2}-(1)^{2}}{(\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}$
$=\frac{\frac{5}{4}+4 \times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}$
$=\frac{\frac{5}{4}+\frac{16}{3}-1}{\frac{1}{4}+\frac{3}{4}}$
$=\frac{15+64-12}{12}$
$=\frac{67}{12}$.