4.2 Given that \( \hat{\mathrm{A}}=38,2^{\circ} \) and \( \hat{\mathrm{B}}=146,4^{\circ} \). Calculate the value of \( 2 \operatorname{cosec} \mathrm{~A}+\cos 3 \mathrm{~B} \).

To calculate \( 2 \operatorname{cosec} \mathrm{~A} + \cos 3 \mathrm{~B} \), we first need to determine the values of \( \operatorname{cosec} \mathrm{~A} \) and \( \cos 3 \mathrm{~B} \). 1. **Calculate \( \operatorname{cosec} \mathrm{~A} \)**: \( \operatorname{cosec} \mathrm{~A} = \frac{1}{\sin \hat{A}} \). First, we find \( \sin \hat{A} \): \[ \hat{A} = 38.2^\circ \implies \sin 38.2^\circ \approx 0.6157. \] Thus, \[ \operatorname{cosec} \mathrm{~A} = \frac{1}{0.6157} \approx 1.622. \] 2. **Calculate \( 2 \operatorname{cosec} \mathrm{~A} \)**: \[ 2 \operatorname{cosec} \mathrm{~A} \approx 2 \times 1.622 \approx 3.244. \] 3. **Calculate \( \cos 3 \mathrm{~B} \)**: We first find \( 3 \hat{B} \): \[ \hat{B} = 146.4^\circ \implies 3 \hat{B} = 3 \times 146.4^\circ = 439.2^\circ. \] Since cosine is periodic with a period of \( 360^\circ \), we can reduce \( 439.2^\circ \): \[ 439.2^\circ - 360^\circ = 79.2^\circ. \] Now, calculate \( \cos 79.2^\circ \): \[ \cos 79.2^\circ \approx 0.1908. \] 4. **Combine the results**: Finally, we add both components: \[ 2 \operatorname{cosec} \mathrm{~A} + \cos 3 \mathrm{~B} \approx 3.244 + 0.1908 \approx 3.4348. \] Thus, the final answer is approximately \( 3.4348 \).