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1

Solved example of special quotients

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\left(\csc\left(x\right)-\cot\left(x\right)\right)^2$
2

A binomial squared (difference) is equal to the square of the first term, minus the double product of the first by the second, plus the square of the second term. In other words: $(a-b)^2=a^2-2ab+b^2$

  • Square of the first term: $\left(\csc\left(x\right)\right)^2 = \csc\left(x\right)^2$
  • Double product of the first by the second: $2\left(\csc\left(x\right)\right)\left(-\cot\left(x\right)\right) = 2\csc\left(x\right)-\cot\left(x\right)$
  • Square of the second term: $\left(-\cot\left(x\right)\right)^2 = \left(-\cot\left(x\right)\right)^2$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\csc\left(x\right)^2-2\csc\left(x\right)\cot\left(x\right)+1\cdot \cot\left(x\right)^2$
3

Any expression multiplied by $1$ is equal to itself

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\csc\left(x\right)^2-2\csc\left(x\right)\cot\left(x\right)+\cot\left(x\right)^2$

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$\frac{1^2}{\sin\left(x\right)^2}$
4

Applying the cosecant identity: $\displaystyle\csc\left(\theta\right)=\frac{1}{\sin\left(\theta\right)}$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{1^2}{\sin\left(x\right)^2}-2\csc\left(x\right)\cot\left(x\right)+\cot\left(x\right)^2$
5

Calculate the power $1^2$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{1}{\sin\left(x\right)^2}-2\csc\left(x\right)\cot\left(x\right)+\cot\left(x\right)^2$
6

Applying the cosecant identity: $\displaystyle\csc\left(\theta\right)=\frac{1}{\sin\left(\theta\right)}$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{1}{\sin\left(x\right)^2}-2\left(\frac{1}{\sin\left(x\right)}\right)\cot\left(x\right)+\cot\left(x\right)^2$
7

Apply the formula: $a\frac{1}{x}$$=\frac{a}{x}$, where $a=-2$ and $x=\sin\left(x\right)$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{1}{\sin\left(x\right)^2}+\frac{-2}{\sin\left(x\right)}\cot\left(x\right)+\cot\left(x\right)^2$
8

Apply the formula: $\cot\left(x\right)$$=\frac{\cos\left(x\right)}{\sin\left(x\right)}$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{1}{\sin\left(x\right)^2}+\frac{-2}{\sin\left(x\right)}\cdot\frac{\cos\left(x\right)}{\sin\left(x\right)}+\cot\left(x\right)^2$
9

Apply the formula: $\cot\left(x\right)$$=\frac{\cos\left(x\right)}{\sin\left(x\right)}$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{1}{\sin\left(x\right)^2}+\frac{-2}{\sin\left(x\right)}\cdot\frac{\cos\left(x\right)}{\sin\left(x\right)}+\left(\frac{\cos\left(x\right)}{\sin\left(x\right)}\right)^2$
10

The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{1}{\sin\left(x\right)^2}+\frac{-2}{\sin\left(x\right)}\cdot\frac{\cos\left(x\right)}{\sin\left(x\right)}+\frac{\cos\left(x\right)^2}{\sin\left(x\right)^2}$
11

Add fraction's numerators with common denominators: $\frac{1}{\sin\left(x\right)^2}$ and $\frac{\cos\left(x\right)^2}{\sin\left(x\right)^2}$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{1+\cos\left(x\right)^2}{\sin\left(x\right)^2}+\frac{-2}{\sin\left(x\right)}\cdot\frac{\cos\left(x\right)}{\sin\left(x\right)}$

When multiplying exponents with same base you can add the exponents

$\sin\left(x\right)^2$
12

Multiplying fractions

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{1+\cos\left(x\right)^2}{\sin\left(x\right)^2}+\frac{-2\cos\left(x\right)}{\sin\left(x\right)^2}$
13

Add fraction's numerators with common denominators: $\frac{1+\cos\left(x\right)^2}{\sin\left(x\right)^2}$ and $\frac{-2\cos\left(x\right)}{\sin\left(x\right)^2}$

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{1+\cos\left(x\right)^2-2\cos\left(x\right)}{\sin\left(x\right)^2}$
14

The trinomial $\frac{1+\cos\left(x\right)^2-2\cos\left(x\right)}{\sin\left(x\right)^2}$ is perfect square, because it's discriminant is equal to zero

$\Delta=b^2-4ac=-2^2-4\left(1\right)\left(1\right) = 0$
15

Using the perfect square trinomial formula

$a^2+2ab+b^2=(a+b)^2,\:where\:a=\sqrt{\cos\left(x\right)^2}\:and\:b=\sqrt{1}$
16

Factoring the perfect square trinomial

$\frac{1-\cos\left(x\right)}{1+\cos\left(x\right)}=\frac{\left(\cos\left(x\right)-1\right)^{2}}{\sin\left(x\right)^2}$
17

Both expressions are not equal

$false$

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