Question

Establish each identity.

$\frac{{(secv-\tan v)}^2+1}{cscv(secv-\tan v)}=2\tan v$

Short answer

The left side of the given equation has been simplified to bring the right side.

Step-by-step solution

Step 1. Given Information

The given equation is$\frac{{(secv-\tan v)}^2+1}{cscv(secv-\tan v)}=2\tan v$

Step 2. Calculation

Consider the left side of the given equation and simplify to bring the right side in order to establish the equation.

$$\begin{gathered} \frac{{(secv-\tan v)}^2+1}{cscv(secv-\tan v)}=\frac{{({\displaystyle \frac{1}{\cos v}}-{\displaystyle \frac{\sin v}{\cos v}})}^2+1}{{\displaystyle \frac{1}{\sin v}}({\displaystyle \frac{1}{\cos v}}-{\displaystyle \frac{\sin v}{\cos v}})} \\ =\frac{{\displaystyle \frac{{(1-\sin v)}^2}{{\cos }^{2}v}}+1}{{\displaystyle \frac{1}{\sin v}}\left({\displaystyle \frac{1-\sin v}{\cos v}}\right)} \\ =\frac{{\displaystyle \frac{{(1-\sin v)}^2+{\cos }^{2}v}{{\cos }^{2}v}}}{{\displaystyle \frac{1-\sin v}{\sin v\cos v}}} \\ =\frac{1+{\sin }^{2}v+{\cos }^{2}v-2\sin v}{\cos v}\times \frac{\sin v}{1-\sin v} \\ =\frac{2-2\sin v}{1-\sin v}\times \frac{\sin v}{\cos v} \\ =\frac{2(1-\sin v)}{1-\sin v}\times \tan v \\ =2\tan v \end{gathered}$$

Hence, Proved.