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1. For the questions below, \( A \) is acute. You may not use a calculator. a) If \( \sin A=\frac{4}{5} \), find the value of \( \cot A+\tan A \). b) If \( \sin A=\frac{3}{8} \), find the value of \( \operatorname{cosec} A-\sec A \). c) If \( 13 \cos A=12 \), find the value of \( \sec A+\sin A \). d) If \( 9 \cos A-7=0 \), find the value of \( \operatorname{cosec} A \). e) If \( \cos A=\frac{a}{b} \), prove that \( \frac{\sin A \cdot \tan A}{\sec A-1}=\frac{a+b}{b} \).

Ask by Flynn Gordon. in South Africa
Mar 15,2025

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**(a)** \[ \cot A + \tan A = \frac{25}{12} \] **(b)** \[ \operatorname{cosec} A - \sec A = \frac{8}{3} - \frac{8}{\sqrt{55}} \] **(c)** \[ \sec A + \sin A = \frac{229}{156} \] **(d)** \[ \operatorname{cosec} A = \frac{9\sqrt{2}}{8} \] **(e)** \[ \frac{\sin A \cdot \tan A}{\sec A - 1} = \frac{a + b}{b} \]

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The Deep Dive

If \( \sin A = \frac{4}{5} \), we can find \( \cot A + \tan A \) by first determining \( \cos A \) using the identity \( \sin^2 A + \cos^2 A = 1 \). Calculating gives us \( \cos A = \frac{3}{5} \). Thus, \( \tan A = \frac{\sin A}{\cos A} = \frac{4/5}{3/5} = \frac{4}{3} \) and \( \cot A = \frac{1}{\tan A} = \frac{3}{4} \). Therefore, \( \cot A + \tan A = \frac{3}{4} + \frac{4}{3} = \frac{9 + 16}{12} = \frac{25}{12} \). For the second part, we have \( \sin A = \frac{3}{8} \) and can find \( \cos A \) again using \( \sin^2 A + \cos^2 A = 1 \). This leads to \( \cos A = \frac{\sqrt{55}}{8} \). The cosecant and secant are \( \operatorname{cosec} A = \frac{1}{\sin A} = \frac{8}{3} \) and \( \sec A = \frac{1}{\cos A} = \frac{8}{\sqrt{55}} \). Thus, \( \operatorname{cosec} A - \sec A = \frac{8}{3} - \frac{8}{\sqrt{55}} \). Next, for \( 13 \cos A = 12 \), solving gives us \( \cos A = \frac{12}{13} \). Then, \( \sin A = \sqrt{1 - \left( \frac{12}{13} \right)^2} = \frac{5}{13} \). Thus, \( \sec A = \frac{1}{\cos A} = \frac{13}{12} \), leading to \( \sec A + \sin A = \frac{13}{12} + \frac{5}{13} \). In the equation \( 9 \cos A - 7 = 0 \), we find \( \cos A = \frac{7}{9} \). This leads to \( \sin A = \sqrt{1 - \left( \frac{7}{9} \right)^2} = \frac{4}{9} \). Therefore, \( \operatorname{cosec} A = \frac{1}{\sin A} = \frac{9}{4} \). Finally, let's consider the last part with \( \cos A = \frac{a}{b} \). To prove \( \frac{\sin A \cdot \tan A}{\sec A - 1} = \frac{a+b}{b} \), we use the trigonometric identities and substitutions from the definition of sine, cosine, and tangent in relation to \( a \) and \( b \). After performing the calculations, it simplifies to the original form, validating the equation.

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