Question

Show$\sqrt{9se{c}^2\theta -9}=3\tan \theta if\mathrm{\pi }\le \mathrm{\theta }<\frac{3\mathrm{\pi }}{2}$

Short answer

It is proved that$\sqrt{9se{c}^2\theta -9}=3\tan \theta$

Step-by-step solution

Step 1. Given Information

The given equation and interval is$\sqrt{9se{c}^2\theta -9}=3\tan \theta ,\mathrm{\pi }\le \mathrm{\theta }<\frac{3\mathrm{\pi }}{2}$

Step 2. Calculation

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

$$\begin{gathered} \sqrt{9se{c}^2\theta -9}=\sqrt{9(se{c}^2\theta -1)} \\ =3\sqrt{se{c}^2\theta -1} \\ =3\sqrt{\frac{1}{{\cos }^2\theta }-1} \\ =3\sqrt{\frac{1-{\cos }^2\theta }{{\cos }^2\theta }} \\ =3\sqrt{\frac{{\sin }^2\theta }{{\cos }^2\theta }} \\ =3\sqrt{{\tan }^2\theta } \\ =3\tan \theta \end{gathered}$$

Hence, Proved