Chapter 7.4, Problem 64AYU

establish each identity.

$\frac{\tan u-\cot u}{\tan u+\cot u}+2{\cos }^{2}u=1$

To determine

To find: The trigonometric expression $\frac{\tan u- \cot u}{\tan u+ \cot u}+2{\text{cos}}^{2}u=1$.

Answer to Problem 64AYU

Solution:

The trigonometric expression $\frac{\tan u- \cot u}{\tan u+ \cot u}+2{\text{cos}}^{2}u=1$ is established.

Explanation of Solution

Given:

The trigonometric expression $\frac{\tan u- \cot u}{\tan u+ \cot u}+2{\text{cos}}^{2}u=1$.

Calculation:

Left hand side,

$\frac{\tan u- \cot u}{\tan u+ \cot u}+2{\text{cos}}^{2}u$

By quotient identity for $\tan u$, $\cot u$.

$\frac{\frac{\sin u}{\cos u}-\frac{\cos u}{\sin u}}{\frac{\sin u}{\cos u}+\frac{\cos u}{\sin u}}+2{\text{cos}}^{2}u$

Adding up the quotients in the numerator and denominator.

$\frac{\frac{{\text{sin}}^{2}u-{\text{ cos}}^{2}u}{\cos u . \sin u}}{\frac{{\text{sin}}^{2}u+ {\text{cos}}^{2}u}{\cos u. \sin u}}+2{\text{cos}}^{2}u$

By Pythagorean identity and cancelling the common factor $\cos u . \sin u$ by reciprocal.

$\frac{{\text{sin}}^{2}u-{\text{ cos}}^{2}u}{1}\text{ + }2{\text{cos}}^{2}u$

By Pythagorean identity,

$(1-{\text{ cos}}^{2}u-{\text{ cos}}^{2}u)+2{\text{cos}}^{2}u$

Open the parenthesis.

$1-2{\text{cos}}^{2}u +2{\text{cos}}^{2}u\text{ = 1}$

$\text{LHS }=\text{ RHS}$