# Question a2c18

Jan 3, 2017

Leave one side unchanged and use algebra and trigonometric substitutions to make the other side the same as the unchanged side.

#### Explanation:

Verify:

cot^2(x)/(sin(x) + cos(x)) =(cos^2(x)sin(x) -cos^3(x))/(2sin^4(x) - sin^2(x))

To verify, I will only change the right side until it is the same as the left side.

Remove common factor of ${\cos}^{2} \left(x\right)$ from the numerator and a common factor ${\sin}^{2} \left(x\right)$ from the denominator:

cot^2(x)/(sin(x) + cos(x)) =(cos^2(x)(sin(x) -cos(x)))/(sin^2(x)(2sin^2(x) - 1)

substitute ${\cot}^{2} \left(x\right)$ into the numerator for ${\cos}^{2} \frac{x}{\sin} ^ 2 \left(x\right)$:

cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/((2sin^2(x) - 1)

Split $2 {\sin}^{2} \left(x\right)$ into ${\sin}^{2} \left(x\right) + {\sin}^{2} \left(x\right)$

cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/((sin^2(x) + sin^2(x) - 1)

Substitute $1 - {\cos}^{2} \left(x\right)$ for the second ${\sin}^{2} \left(x\right)$:

cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/((sin^2(x) + 1 - cos^2(x) - 1)

The 1s cancel:

cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/((sin^2(x) cancel(+ 1) - cos^2(x) cancel(- 1))

cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/((sin^2(x) - cos^2(x))

The denominator is the difference of two squares and we know how that factors:

cot^2(x)/(sin(x) + cos(x)) =(cot^2(x)(sin(x) -cos(x)))/(((sin(x) - cos(x))(sin(x) + cos(x)))#

The $\frac{\sin \left(x\right) - \cos \left(x\right)}{\sin \left(x\right) - \cos \left(x\right)}$ cancels:

${\cot}^{2} \frac{x}{\sin \left(x\right) + \cos \left(x\right)} = \frac{{\cot}^{2} \left(x\right) \cancel{\sin \left(x\right) - \cos \left(x\right)}}{\left(\cancel{\sin \left(x\right) - \cos \left(x\right)}\right) \left(\sin \left(x\right) + \cos \left(x\right)\right)}$

${\cot}^{2} \frac{x}{\sin \left(x\right) + \cos \left(x\right)} = {\cot}^{2} \frac{x}{\sin \left(x\right) + \cos \left(x\right)}$

The right side is the same as the left side. Q.E.D.