Question

If 15 cot N = 8, find the value of $\frac{1}{\sqrt{se{c}^{2}N-1}}$

Answer 1

Given:

$$\begin{gathered} 15\cot \mathrm{N}=8 \\ \Rightarrow \cot \mathrm{N}=\frac{8}{15} \end{gathered}$$
We know that $\cot \mathrm{N}=\frac{\mathrm{base}}{\mathrm{perpendicular}}=\frac{8}{15}$
Let base = 8k and perpendicular = 15k, where k is a positive constant
In the right triangle, we have:
$$\begin{gathered} {\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{perpendicular}\right)}^2+{\left(\mathrm{base}\right)}^2(\mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}) \\ \Rightarrow {\left(\mathrm{Hypotenuse}\right)}^2={\left(15k\right)}^2+{\left(8k\right)}^2 \\ \Rightarrow {\left(\mathrm{Hypotenuse}\right)}^2=225k^2+64k^2 \\ \Rightarrow {\left(\mathrm{Hypotenuse}\right)}^2=289k^2 \\ \Rightarrow \left(\mathrm{Hypotenuse}\right)=\pm 17k \\ \Rightarrow {\left(\mathrm{Hypotenuse}\right)}^2=17k(\because \mathrm{Length}\mathrm{cannot}\mathrm{be}\mathrm{negative}) \\ \therefore \sec \mathrm{N}=\frac{\mathrm{hypotenuse}}{\mathrm{base}}=\frac{17}{8} \end{gathered}$$


$$\begin{gathered} \mathrm{Now},\mathrm{consider}\frac{1}{\sqrt{se{c}^{2}N-1}} \\ =\frac{1}{\sqrt{{\left({\displaystyle \frac{17}{8}}\right)}^2-1}} \\ =\frac{1}{\sqrt{{\displaystyle \frac{289}{64}}-1}} \\ =\sqrt{\frac{64}{225}} \\ =\frac{8}{15} \end{gathered}$$