Chapter 7.4, Problem 77AYU

establish each identity.

$\frac{{\left(\sec \nu -\tan \nu \right)}^2+1}{\csc \nu \left(\sec \nu -\tan \nu \right)}=2\tan \nu$

To determine

To find: The trigonometric expression $\frac{{\left(\sec v-\tan v\right)}^2+1}{\csc v\left(\sec v-\tan v\right)}=2\tan v$.

Answer to Problem 77AYU

Solution:

The trigonometric expression $\frac{{\left(\sec v-\tan v\right)}^2+1}{\csc v\left(\sec v-\tan v\right)}=2\tan v$ is established.

Explanation of Solution

Given:

The trigonometric expression $\frac{{\left(\sec v-\tan v\right)}^2+1}{\csc v\left(\sec v-\tan v\right)}=2\tan v$.

Calculation:

Left hand side

$\frac{{\left(\sec v-\tan v\right)}^2+1}{\csc v\left(\sec v-\tan v\right)}$

Expanding numerator

$\frac{({\text{sec}}^{2}v+{\text{ tan}}^{2}v-2\sec v\tan v)+1}{\csc v\left(\sec v-\tan v\right)}$

By Quotient identity

$\frac{(\frac{1}{{\text{cos}}^2\theta }+\frac{{\text{sin}}^2\theta }{{\text{cos}}^2\theta }-2\frac{1}{\cos v}\frac{\sin v}{\cos v})+1}{\frac{1}{\sin v}\left(\frac{1}{\cos v}-\frac{\sin v}{\cos v}\right)}$

Adding up the quotients in the numerator and denominator

$\frac{\frac{1+{\text{ sin}}^{2}v-2\sin v}{{\text{cos}}^{2}v}+1}{\frac{1}{\sin v}\left(\frac{1-\sin v}{\cos v}\right)}$

Cancelling the common factor $\cos v$by reciprocal

$\frac{\frac{1+{\text{ sin}}^{2}v-2\sin v+{\text{ cos}}^2\theta }{\cos v}}{\frac{1}{\sin v}\left(1-\sin v\right)}$

By Pythagorean identity

$\frac{\frac{\left(1+1-2\sin v\right).\sin v}{\cos v}}{\left(1-\sin v\right)}$

$\frac{\frac{2-2\sin v}{\cos v}.\sin v}{\left(1-\sin v\right)}$

Taking the common factor 2 from the common factor

$\frac{\frac{2\left(1-\sin v\right)}{\cos v}.\sin v}{\left(1-\sin v\right)}$

By quotient identity

$\frac{2\left(1-\sin v\right).\tan v}{\left(1-\sin v\right)}$

Cancelling the common factor $(1-\sin v)$

$2\tan v$

$\text{LHS = RHS}$