Evaluate the following : (i) sin 60° cos 30° + sin 30° cos 60° (ii) 2 tan2 45° + cos2 30° – sin2 60° (iii) cos 45° sec 30° + cosec 30° (iv) sin 30° + tan 45° – cosec 60° sec 30° + cos 60° + cot 45° (v) 2 2 2 2 2 5 cos 60 4 sec 30 tan 45 sin 30 cos 30
Step-by-Step Solution
Key Concept: Use the standard trigonometric values for 30°, 45°, 60° and the fundamental identities: <br>- $<br>sin(A+B)=sinA\,cosB+cosA\,sinB$<br>- $<br>cos^2\theta+sin^2\theta=1$<br>- Reciprocal definitions $\sec\theta=1/\cos\theta$, $\cosec\theta=1/\sin\theta$, $\cot\theta=1/\tan\theta$.<br>All calculations are performed by substituting the exact values $\sin30°=\frac12$, $\cos30°=\frac{\sqrt3}{2}$, $\sin45°=\cos45°=\frac{\sqrt2}{2}$, $\sin60°=\cos60°=\frac{\sqrt3}{2}$, $\tan45°=1$, etc.
### (i) $\sin 60^{\circ}\cos 30^{\circ}+\sin 30^{\circ}\cos 60^{\circ}$
Using the identity $\sin(A+B)=\sin A\cos B+\cos A\sin B$,
$$\sin 60^{\circ}\cos 30^{\circ}+\sin 30^{\circ}\cos 60^{\circ}=\sin(60^{\circ}+30^{\circ})=\sin 90^{\circ}=1.$$
Alternatively, substituting the exact values:
$$\sin60^{\circ}=\frac{\sqrt3}{2},\;\cos30^{\circ}=\frac{\sqrt3}{2},\;\sin30^{\circ}=\frac12,\;\cos60^{\circ}=\frac12$$
$$\Rightarrow \frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2}+\frac12\cdot\frac12=\frac{3}{4}+\frac{1}{4}=1.$$
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### (ii) $2\tan^{2}45^{\circ}+\cos^{2}30^{\circ}-\sin^{2}60^{\circ}$
\[\tan45^{\circ}=1\Rightarrow\tan^{2}45^{\circ}=1\]
\[\cos30^{\circ}=\frac{\sqrt3}{2}\Rightarrow\cos^{2}30^{\circ}=\frac{3}{4}\]
\[\sin60^{\circ}=\frac{\sqrt3}{2}\Rightarrow\sin^{2}60^{\circ}=\frac{3}{4}\]
Hence
$$2\times1+\frac{3}{4}-\frac{3}{4}=2.$$
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### (iii) $\cos45^{\circ}\sec30^{\circ}+\cosec30^{\circ}$
\[\cos45^{\circ}=\frac{\sqrt2}{2},\qquad \sec30^{\circ}=\frac{1}{\cos30^{\circ}}=\frac{2}{\sqrt3}\]
$$\cos45^{\circ}\sec30^{\circ}=\frac{\sqrt2}{2}\cdot\frac{2}{\sqrt3}=\frac{\sqrt2}{\sqrt3}=\frac{\sqrt6}{3}$$
\[\cosec30^{\circ}=\frac{1}{\sin30^{\circ}}=\frac{1}{\frac12}=2\]
Therefore
$$\cos45^{\circ}\sec30^{\circ}+\cosec30^{\circ}=\frac{\sqrt6}{3}+2=\frac{6+\sqrt6}{3}.$$
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### (iv) $\sin30^{\circ}+\tan45^{\circ}-\cosec60^{\circ}\sec30^{\circ}+\cos60^{\circ}+\cot45^{\circ}$
Values:
\[\sin30^{\circ}=\frac12,\;\tan45^{\circ}=1,\;\cosec60^{\circ}=\frac{2}{\sqrt3},\;\sec30^{\circ}=\frac{2}{\sqrt3},\;\cos60^{\circ}=\frac12,\;\cot45^{\circ}=1\]
Compute the product:
$$\cosec60^{\circ}\sec30^{\circ}=\frac{2}{\sqrt3}\cdot\frac{2}{\sqrt3}=\frac{4}{3}$$
Now sum all terms:
$$\frac12+1-\frac{4}{3}+\frac12+1=\left(\frac12+\frac12\right)+(1+1)-\frac{4}{3}=1+2-\frac{4}{3}=\frac{5}{3}.$$
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### (v) $2\cos60^{\circ}+2\sec30^{\circ}-2\tan45^{\circ}+2\sin30^{\circ}+2\cos30^{\circ}$
Substituting the standard values:
\[2\cos60^{\circ}=2\times\frac12=1\]
\[2\sec30^{\circ}=2\times\frac{2}{\sqrt3}=\frac{4}{\sqrt3}=\frac{4\sqrt3}{3}\]
\[2\tan45^{\circ}=2\times1=2\]
\[2\sin30^{\circ}=2\times\frac12=1\]
\[2\cos30^{\circ}=2\times\frac{\sqrt3}{2}=\sqrt3\]
Adding them:
$$1+\frac{4\sqrt3}{3}-2+1+\sqrt3=\left(1+1-2\right)+\left(\frac{4\sqrt3}{3}+\sqrt3\right)=0+\frac{4\sqrt3+3\sqrt3}{3}=\frac{7\sqrt3}{3}.$$
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Final Answers
- (i) $1$
- (ii) $2$
- (iii) $\dfrac{6+\sqrt6}{3}$
- (iv) $\dfrac{5}{3}$
- (v) $\dfrac{7\sqrt3}{3}$
Correct Answer: (i) 1, (ii) 2, (iii) (6+√6)/3, (iv) 5/3, (v) 7√3/3