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QUESTION 3 Determine the value of \( \theta \) if \( \theta \in\left[0^{\circ} ; 90^{\circ}\right] \), without the use of a calculator. \( \frac{\tan 30^{\circ \circ} \cdot \operatorname{cosec} 60^{\circ}}{\cos 45^{\circ} \cdot \sin 45^{\circ}} \)

Ask by Mullins Herrera. in South Africa
Mar 09,2025

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Answer

\( \theta = \tan^{-1}\left(\frac{4}{3}\right) \)

Solution

To determine the value of \( \theta \) from the expression \[ \frac{\tan 30^{\circ} \cdot \operatorname{cosec} 60^{\circ}}{\cos 45^{\circ} \cdot \sin 45^{\circ}}, \] we will first evaluate each trigonometric function involved in the expression. 1. **Calculate \( \tan 30^{\circ} \)**: \[ \tan 30^{\circ} = \frac{1}{\sqrt{3}}. \] 2. **Calculate \( \operatorname{cosec} 60^{\circ} \)**: \[ \operatorname{cosec} 60^{\circ} = \frac{1}{\sin 60^{\circ}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}. \] 3. **Calculate \( \cos 45^{\circ} \) and \( \sin 45^{\circ} \)**: \[ \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2}. \] Now, substitute these values into the expression: \[ \frac{\tan 30^{\circ} \cdot \operatorname{cosec} 60^{\circ}}{\cos 45^{\circ} \cdot \sin 45^{\circ}} = \frac{\left(\frac{1}{\sqrt{3}}\right) \cdot \left(\frac{2}{\sqrt{3}}\right)}{\left(\frac{\sqrt{2}}{2}\right) \cdot \left(\frac{\sqrt{2}}{2}\right)}. \] 4. **Simplify the numerator**: \[ \text{Numerator} = \frac{1 \cdot 2}{\sqrt{3} \cdot \sqrt{3}} = \frac{2}{3}. \] 5. **Simplify the denominator**: \[ \text{Denominator} = \frac{\sqrt{2} \cdot \sqrt{2}}{2 \cdot 2} = \frac{2}{4} = \frac{1}{2}. \] Now, we can rewrite the entire expression: \[ \frac{\frac{2}{3}}{\frac{1}{2}} = \frac{2}{3} \cdot 2 = \frac{4}{3}. \] Now we have: \[ \frac{4}{3} = \tan \theta. \] To find \( \theta \), we need to determine the angle whose tangent is \( \frac{4}{3} \). Since \( \theta \) is in the range \( [0^{\circ}, 90^{\circ}] \), we can conclude that: \[ \theta = \tan^{-1}\left(\frac{4}{3}\right). \] Thus, the value of \( \theta \) is: \[ \theta = \tan^{-1}\left(\frac{4}{3}\right). \]

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Beyond the Answer

To evaluate the expression \( \frac{\tan 30^{\circ} \cdot \operatorname{cosec} 60^{\circ}}{\cos 45^{\circ} \cdot \sin 45^{\circ}} \), let's first recall some trigonometric values: 1. \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \) 2. \( \operatorname{cosec} 60^{\circ} = \frac{1}{\sin 60^{\circ}} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \) 3. \( \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2} \) Now substituting these values into the expression: \[ \frac{\tan 30^{\circ} \cdot \operatorname{cosec} 60^{\circ}}{\cos 45^{\circ} \cdot \sin 45^{\circ}} = \frac{\left(\frac{1}{\sqrt{3}}\right) \cdot \left(\frac{2}{\sqrt{3}}\right)}{\left(\frac{\sqrt{2}}{2}\right) \cdot \left(\frac{\sqrt{2}}{2}\right)} \] Calculating the numerator: \[ \frac{1 \cdot 2}{\sqrt{3} \cdot \sqrt{3}} = \frac{2}{3} \] Calculating the denominator: \[ \frac{\sqrt{2} \cdot \sqrt{2}}{2 \cdot 2} = \frac{2}{4} = \frac{1}{2} \] Putting it all together: \[ \frac{\frac{2}{3}}{\frac{1}{2}} = \frac{2}{3} \cdot \frac{2}{1} = \frac{4}{3} \] To find \( \theta \) such that \( \theta \in [0^{\circ}, 90^{\circ}] \) and \( \tan \theta = \frac{4}{3} \), we can identify the angle that corresponds to this ratio. This leads us to conclude that: \[ \theta = \tan^{-1} \left(\frac{4}{3}\right) \] Hence, the value of \( \theta \) can be determined as \( \theta \) which lies between \( 0^{\circ} \) and \( 90^{\circ} \).

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