Question

A searchlight at the grand opening of a new car dealership casts a spot of light on a wall located 75 meters from the searchlight. The acceleration r of the spot of light is found to be $r=1200sec\theta (2se{c}^2\theta -1)$. Show that this is equivalent to$r=1200\left(\frac{1+{\sin }^2\theta }{{\cos }^3\theta }\right)$

Short answer

The given expressions are equivalent.

Step-by-step solution

Step 1. Given Information

The given expressions are$r=1200sec\theta (2se{c}^2\theta -1)andr=1200\left(\frac{1+{\sin }^2\theta }{{\cos }^3\theta }\right)$

Step 2. Calculation

Consider one expression and prove it is equivalent to the other given expression.

$$\begin{gathered} 1200sec\theta (2se{c}^2\theta -1)=\frac{1200}{\cos \theta }(\frac{2}{{\cos }^2\theta }-1) \\ =\frac{1200}{\cos \theta }\left(\frac{2-{\cos }^2\theta }{{\cos }^2\theta }\right) \\ =\frac{1200}{\cos \theta }\left(\frac{1+1-{\cos }^2\theta }{{\cos }^2\theta }\right) \\ =\frac{1200}{\cos \theta }\left(\frac{1+{\sin }^2\theta }{{\cos }^2\theta }\right) \\ =\frac{1200}{{\cos }^3\theta }(1+{\sin }^2\theta ) \\ =r \end{gathered}$$

Thus, we can conclude that the given expressions are equivalent.