Question : If $\cot B=\frac{15}{8}$, where $B$ is an acute angle, what is the value of $\sec B+\tan B$?
Option 1: $\frac{4}{5}$
Option 2: $\frac{5}{3}$
Option 3: $\frac{5}{4}$
Option 4: $\frac{3}{5}$
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Correct Answer: $\frac{5}{3}$
Solution : Given: $\cot B=\frac{15}{8}$ We know that $\cot B=\frac{\text{base (b)}}{\text{perpendicular (p)}}=\frac{15}{8}$ So, $h^2 = \sqrt{b^2+p^2}$ where $h$ is hypotenuse. ⇒ $h^2 = \sqrt{15^2+8^2}$ ⇒ $h^2 = \sqrt{225+64}$ ⇒ $h^2 = \sqrt{289}$ ⇒ $h = 17$ Thus, $\sec B+\tan B$ $= \frac{\text{h}}{\text{b}}+\frac{\text{p}}{\text{b}}$ Putting the values of $p, b$, and $h$, we get $= \frac{17}{15}+\frac{8}{15}$ $= \frac{25}{15}$ $= \frac{5}{3}$ Hence, the correct answer is $\frac{5}{3}$.
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