Solved example of special quotients

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: $\left(-\cot\left(x\right)\right)^2 = \left(-\cot\left(x\right)\right)^2$

Any expression multiplied by $1$ is equal to itself

Calculate the power $1^2$

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

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

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

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

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

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

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

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

Using the perfect square trinomial formula

Factoring the perfect square trinomial

Both expressions are not equal