If $\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$, show that $\csc x \cot x=\frac{a^2-b^2}{4ab}$

Question

If $$\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$$ show that $$\csc x \cot x=\frac{a^2-b^2}{4ab}$$

(original problem image)

I tried getting rid of the denominator and then expanding the given equation, but couldn't get to the answer.

Answer 1

Suppose $\sin x \neq 0$. Then we have that $\csc x = \dfrac{1}{\sin x}$ and $\dfrac{\cos x}{\sin x} = \cot x$

This implies in $\cos x = \dfrac{a-b}{a+b}$. Therefore, $\csc x \cdot \cot x = \dfrac {\cos x}{\sin^2 x} = \dfrac {\cos x}{1-\cos^2 x}.$

Notice that $1- \cos^2 x = 1 - \left(\dfrac{a-b}{a+b}\right)^2 = \dfrac{4ab}{(a+b)^2}$.

Then, $\dfrac {\cos x}{1-\cos^2 x}=\dfrac{(a-b)(a+b)}{4ab}=\dfrac{a^2-b^2}{4ab}.$

Answer 2

From $1+\cot^2 x= \csc^2 x$ we have \begin{eqnarray*} (\color{red}{\csc x} +\cot x)(\csc x -\color{blue}{\cot x}) = 1 \end{eqnarray*} Now sub \begin{eqnarray*} \color{red}{\csc x} = \frac{a+b}{a-b} \cot x \end{eqnarray*} and \begin{eqnarray*} \color{blue}{\cot x} = \frac{a-b}{a+b} \csc x . \end{eqnarray*} This gives \begin{eqnarray*} \cot x \left( \frac{a+b}{a-b} +1 \right) \csc x \left( 1-\frac{a-b}{a+b}\right) =1 \end{eqnarray*} and the result follows with a bit of algebra.

Answer 3

$$ \frac{a+b}{\csc x}=\frac{a-b}{\cot x}=k \qquad\implies\qquad \begin{align} a+b&=k\csc x \\ a-b&=k\cot x \end{align}$$ Therefore, $$\begin{align} a^2 - b^2 &= ( a+ b )( a - b ) &&= k^2 \cdot \csc x \cot x \\ 4 a b &= ( a + b )^2 - ( a - b )^2 = k^2 (\csc^2x-\cot^2x)&&=k^2\cdot 1 \end{align}$$ The result follows. $\square$

Answer 4

Rearrange

$$\frac{a^2-b^2}{4ab}=\frac{(a+b)(a-b)}{(a+b)^2-(a-b)^2} =\frac{1}{\frac{a+b}{a-b}-\frac{a-b}{a+b}}$$

Then, substitute $\frac{a+b}{a-b}=\frac{\csc x}{\cot x}$,

$$\frac{a^2-b^2}{4ab}=\frac1{\frac{\csc x}{\cot x}-\frac{\cot x}{\csc x}} =\frac{\csc x\cot x}{\csc^2 x-\cot^2x}=\csc x\cot x$$

Answer 5

Like the formulas mentioned here in https://www.onlinemath4all.com/properties-of-proportion.html,

$$\dfrac{\csc x}{a+b}=\dfrac{\cot x}{a-b}=p\sqrt{\dfrac{\csc^2x-\cot^2x}{(a+b)^2-(a-b)^2}}$$

where $p=\pm1\implies p^2=1$