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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