Question

Show$\sqrt{16+16{\tan }^2\theta }=4sec\theta if\frac{-\mathrm{\pi }}{2}<\theta <\frac{\mathrm{\pi }}{2}$

Short answer

It is proved that$\sqrt{16+16{\tan }^2\theta }=4sec\theta$

Step-by-step solution

Step 1. Given Information

The given equation and interval is$\sqrt{16+16{\tan }^2\theta }=4sec\theta ,\frac{-\mathrm{\pi }}{2}<\theta <\frac{\mathrm{\pi }}{2}$

Step 2. Calculation

Consider the left hand side of the equation and prove it to the right side of the given equation.

$$\begin{gathered} \sqrt{16+16{\tan }^2\theta }=\sqrt{16(1+{\tan }^2\theta )} \\ =4\sqrt{1+{\tan }^2\theta } \\ =4\sqrt{1+\frac{{\sin }^2\theta }{{\cos }^2\theta }} \\ =4\sqrt{\frac{{\cos }^2\theta +{\sin }^2\theta }{{\cos }^2\theta }} \\ =4\sqrt{\frac{1}{{\cos }^2\theta }} \\ =4\sqrt{se{c}^2\theta } \\ =4sec\theta \end{gathered}$$

Hence, proved