If same earns $2 more, he’ll have $24. How much does

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byjennjenn

02.12.2021

$\dfrac{1+\sec a}{\sec a-1} + \dfrac{1+\cos a}{\cos a-1}= 0$ we can change $\frac{1+\sec a}{\sec a-1}$ into cosines. by definition, $\sec a = \frac{1}{\cos a}$ $\begin{aligned} \dfrac{1+\sec a}{\sec a-1} + \dfrac{1+\cos a}{\cos a-1} & = 0 \\ \dfrac{1+\frac{1}{\cos a}}{\frac{1}{\cos a}-1} + \dfrac{1+\cos a}{\cos a-1} & = 0 \end{aligned}$ multiply the numerator and denominator of $\frac{1+\frac{1}{\cos a}}{\frac{1}{\cos a}-1}$ by $\cos a$ so that it can distribute and cancel out those tiny denominators. $\begin{aligned} \dfrac{\cos a\left(1+\frac{1}{\cos a}\right)}{\cos a\left(\frac{1}{\cos a}-1\right)} + \dfrac{1+\cos a}{\cos a-1} & = 0 \\ \dfrac{\cos a+1}{1 - \cos a} + \dfrac{1+\cos a}{\cos a-1} & = 0 \\ \dfrac{\cos a+1}{\boxed{1 - \cos a}} + \dfrac{1+\cos a}{\cos a-1} & = 0 \end{aligned}$ if we want those two fractions to have the same denominator to allow for fraction arithmetic, factor out a negative from $1 - \cos a$ $\begin{aligned} 1 - \cos a & = + \cos a) \\ & = -(\cos a - 1) \end{aligned}$ hence $\begin{aligned} \dfrac{\cos a+1}{1 - \cos a} + \dfrac{1+\cos a}{\cos a-1} & = 0 \\ \dfrac{\cos a+1}{-(\cos a -1)} + \dfrac{1+\cos a}{\cos a-1} & = 0 \\ \dfrac{-(\cos a+1)}{\cos a -1} + \dfrac{1+\cos a}{\cos a-1} & = 0 \\ \dfrac{-(\cos a+1) + 1+\cos a}{\cos a -1} & = 0 \\ \dfrac{-\cos a-1 + 1+\cos a}{\cos a -1} & = 0 \\ \dfrac{\cos a-\cos a + 1 - 1}{\cos a -1} & = 0 \\ \dfrac{0}{\cos a -1} & = 0 \\ 0 & = 0 \end{aligned}$

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If same earns $2 more, he’ll have $24. How much does he have to start out with?

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