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}} $.

AcademicMathematicsNCERTClass 10

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,

$\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}= \frac{5(\frac{1}{2})^{2}+4(\frac{2}{\sqrt{3}})^{2}-(1)^{2}}{(\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}$

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

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

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

$=\frac{67}{12}$.

raja
Updated on 10-Oct-2022 13:22:12

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