Use the special angles to evaluate: a) $$ \operatorname{cosec} 60^{\circ} \cdot \cot 30^{\circ}+\cos 45^{\circ} \cdot \operatorname{cosec} 45^{\circ} $$ b) $$ \sec ^{2} 45^{\circ}- an ^{2} 45^{\circ} $$ c) $$ 1+\cot ^{2} 60^{\circ} $$ d) $$ \operatorname{cosec}^{2} 60^{\circ}+\cot ^{2} 45^{\circ}+\sec ^{2} 30^{\circ} $$ e) $$ \sin 90^{\circ}-\cos 60^{\circ} \cdot \sec 60^{\circ}+\operatorname{cosec} 45^{\circ} \cdot \cos 45^{\circ} $$ f) $$ \operatorname{cosec} 60^{\circ} \cdot an 45^{\circ} \cdot \sin 30^{\circ} $$

Question

Use the special angles to evaluate: a) $$ \operatorname{cosec} 60^{\circ} \cdot \cot 30^{\circ}+\cos 45^{\circ} \cdot \operatorname{cosec} 45^{\circ} $$ b) $$ \sec ^{2} 45^{\circ}-	an ^{2} 45^{\circ} $$ c) $$ 1+\cot ^{2} 60^{\circ} $$ d) $$ \operatorname{cosec}^{2} 60^{\circ}+\cot ^{2} 45^{\circ}+\sec ^{2} 30^{\circ} $$ e) $$ \sin 90^{\circ}-\cos 60^{\circ} \cdot \sec 60^{\circ}+\operatorname{cosec} 45^{\circ} \cdot \cos 45^{\circ} $$ f) $$ \operatorname{cosec} 60^{\circ} \cdot 	an 45^{\circ} \cdot \sin 30^{\circ} $$

UpStudy AI · Accepted Answer

a) We start with  
$$
\csc 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}, \quad \cot 30^\circ = \frac{\cos 30^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}.
$$  
Then,  
$$
\csc 60^\circ \cdot \cot 30^\circ = \frac{2}{\sqrt{3}} \cdot \sqrt{3} = 2.
$$

Next, we have  
$$
\cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}.
$$  
Thus,  
$$
\cos 45^\circ \cdot \csc 45^\circ = \frac{\sqrt{2}}{2} \cdot \sqrt{2} = \frac{2}{2} = 1.
$$

Adding the two parts,  
$$
\csc 60^\circ \cdot \cot 30^\circ + \cos 45^\circ \cdot \csc 45^\circ = 2 + 1 = 3.
$$

---

b) We know that  
$$
\tan 45^\circ = 1, \quad \sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}.
$$  
Thus,  
$$
\sec^2 45^\circ = (\sqrt{2})^2 = 2, \quad \tan^2 45^\circ = 1.
$$

Therefore,  
$$
\sec^2 45^\circ - \tan^2 45^\circ = 2 - 1 = 1.
$$

---

c) We have  
$$
\cot 60^\circ = \frac{\cos 60^\circ}{\sin 60^\circ} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}},
$$  
so  
$$
\cot^2 60^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}.
$$

Thus,  
$$
1 + \cot^2 60^\circ = 1 + \frac{1}{3} = \frac{4}{3}.
$$

---

d) Compute each term separately:

1. For $$ \csc^2 60^\circ $$:
   $$
   \csc 60^\circ = \frac{1}{\sin 60^\circ} = \frac{2}{\sqrt{3}}, \quad \csc^2 60^\circ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}.
   $$
2. For $$ \cot^2 45^\circ $$:
   $$
   \cot 45^\circ = 1 \quad \Rightarrow \quad \cot^2 45^\circ = 1.
   $$
3. For $$ \sec^2 30^\circ $$:
   $$
   \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sec 30^\circ = \frac{1}{\cos 30^\circ} = \frac{2}{\sqrt{3}}, \quad \sec^2 30^\circ = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}.
   $$

Now, adding all three terms:
$$
\csc^2 60^\circ + \cot^2 45^\circ + \sec^2 30^\circ = \frac{4}{3} + 1 + \frac{4}{3} = 1 + \frac{8}{3} = \frac{11}{3}.
$$

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e) Evaluate each term:

1. $$\sin 90^\circ = 1.$$
2. For $$\cos 60^\circ \cdot \sec 60^\circ$$:
   $$
   \cos 60^\circ = \frac{1}{2}, \quad \sec 60^\circ = \frac{1}{\cos 60^\circ} = 2, \quad \cos 60^\circ \cdot \sec 60^\circ = \frac{1}{2} \cdot 2 = 1.
   $$
3. For $$\csc 45^\circ \cdot \cos 45^\circ$$:
   $$
   \csc 45^\circ = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \csc 45^\circ \cdot \cos 45^\circ = \sqrt{2} \cdot \frac{\sqrt{2}}{2} = \frac{2}{2} = 1.
   $$

Thus,  
$$
\sin 90^\circ - \cos 60^\circ \cdot \sec 60^\circ + \csc 45^\circ \cdot \cos 45^\circ = 1 - 1 + 1 = 1.
$$

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f) Compute each factor:

1. $$\csc 60^\circ = \frac{1}{\sin 60^\circ} = \frac{2}{\sqrt{3}}.$$
2. $$\tan 45^\circ = 1.$$
3. $$\sin 30^\circ = \frac{1}{2}.$$

Multiplying the three factors:
$$
\csc 60^\circ \cdot \tan 45^\circ \cdot \sin 30^\circ = \frac{2}{\sqrt{3}} \cdot 1 \cdot \frac{1}{2} = \frac{1}{\sqrt{3}}.
$$
a) 3 b) 1 c) $$\frac{4}{3}$$ d) $$\frac{11}{3}$$ e) 1 f) $$\frac{1}{\sqrt{3}}$$

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