A trigonometric identities with the ratio of four terms like $1+(\frac{\tan x}{\sin y})^2$ Prove:
$$\frac{1+\left(\frac{\tan x}{\sin y}\right)^2}{1+\left(\frac{\tan x}{\sin z}\right)^2}=\frac{1+\left(\frac{\sin x}{\tan y}\right)^2}{1+\left(\frac{\sin x}{\tan z}\right)^2}$$
I started by opening the brackets and squaring but did not get the required answer.
 A: The trigonometric functions for each angle can be rewritten as:
$$
\begin{eqnarray}
\sin^2 x &=& \frac{\alpha}{\alpha + \beta} &\quad \tan^2 x &=& \frac{\alpha}{\beta} \\
\sin^2 y &=& \frac{\gamma}{\gamma + \delta} &\quad \tan^2 y &=& \frac{\gamma}{\delta} \\
\sin^2 z &=& \frac{\epsilon}{\epsilon + \zeta} &\quad \tan^2 z &=& \frac{\epsilon}{\zeta} \\
\end{eqnarray}
$$
The trigonometric identity can then be rewritten as:
$$
\frac{1 + \frac{\alpha (\gamma + \delta)}{\beta \gamma}}{1 + \frac{\alpha (\epsilon + \zeta)}{\beta \epsilon}}
\stackrel{\text{?}}{=} \frac{1 + \frac{\alpha \delta}{\gamma (\alpha + \beta)}}{1 + \frac{\alpha \zeta}{\epsilon (\alpha + \beta)}} \\
$$
After expanding the fractions within fractions, we have:
$$
\frac{\beta \epsilon (\beta \gamma + \alpha (\gamma + \delta))}{\beta \gamma (\beta \epsilon + \alpha (\epsilon + \zeta))}
\stackrel{\text{?}}{=} \frac{\epsilon (\alpha + \beta) (\gamma (\alpha + \beta) + \alpha \delta)}{\gamma (\alpha + \beta) (\epsilon (\alpha + \beta) + \alpha \zeta)} \\
$$
After cancelling the obvious factors, we have:
$$
\frac{\beta \gamma + \alpha (\gamma + \delta)}{\beta \epsilon + \alpha (\epsilon + \zeta)}
\stackrel{\text{?}}{=} \frac{\gamma (\alpha + \beta) + \alpha \delta}{\epsilon (\alpha + \beta) + \alpha \zeta} \\
$$
After expanding the parentheses and ordering to be soft on the eyes, we have:
$$
\frac{\alpha \delta + \alpha \gamma + \beta \gamma}{\alpha \epsilon + \alpha \zeta + \beta \epsilon}
= \frac{\alpha \delta + \alpha \gamma + \beta \gamma}{\alpha \epsilon + \alpha \zeta + \beta \epsilon} \\
$$
Q.E.D.
A: $$\frac{1+\left(\dfrac{\tan x}{\sin y}\right)^2}{1+\left(\dfrac{\tan x}{\sin z}\right)^2}=\frac{1+\left(\dfrac{\sin x}{\tan y}\right)^2}{1+\left(\dfrac{\sin x}{\tan z}\right)^2}$$
$$\iff\frac{1+\left(\dfrac{\tan x}{\sin y}\right)^2}{1+\left(\dfrac{\sin x}{\tan y}\right)^2}=\frac{1+\left(\dfrac{\tan x}{\sin z}\right)^2}{1+\left(\dfrac{\sin x}{\tan z}\right)^2}$$
So, if we can prove that $\dfrac{1+\left(\dfrac{\tan x}{\sin A}\right)^2}{1+\left(\dfrac{\sin x}{\tan A}\right)^2}$ is independent of $A$, we are done.
Method $\#1:$
$\dfrac{1+\left(\dfrac{\tan x}{\sin A}\right)^2}{1+\left(\dfrac{\sin x}{\tan A}\right)^2}=\dfrac{1+\tan^2x\csc^2A}{1+\sin^2x\cot^2A}$
$=\dfrac{\cos^2A+\sin^2x(1+\cot^2A)}{\cos^2x(1+\sin^2x\cot^2A)}=\sec^2x$ which is clearly independent of $A$
Method $\#2:$
$\dfrac{1+\left(\dfrac{\tan x}{\sin A}\right)^2}{1+\left(\dfrac{\sin x}{\tan A}\right)^2}=\dfrac{(\sin^2A\cos^2x+\sin^2x)\sin^2A}{\sin^2A\cos^2x(\sin^2A+\cos^2A\sin^2x)}=\sec^2x$ as $\sin^2A\cos^2x+\sin^2x=\sin^2A(1-\sin^2x)+\sin^2x=\sin^2A+\sin^2x(1-\sin^2A)=?$
A: Notice, the given equality can be easily proved by simplifying $LHS$  $$LHS=\frac{1+\left(\frac{\tan x}{\sin y}\right)^2}{1+\left(\frac{\tan x}{\sin z}\right)^2}$$
$$=\frac{1+\left(\frac{\sin x}{\sin y\cos x}\right)^2}{1+\left(\frac{\sin x}{\sin z\cos x}\right)^2}$$
$$=\frac{\sin^2z(\sin^2 y\cos^2 x+\sin^2 x)}{\sin^2y(\sin^2 z\cos^2 x+\sin^2 x)}$$
$$=\frac{\sin^2z((1-\cos^2 y)(1-\sin^2 x)+\sin^2 x)}{\sin^2y((1-\cos^2 z)(1-\sin^2 x)+\sin^2 x)}$$
$$=\frac{\sin^2z(1-\cos^2 y-\sin^2 x+\sin^2 x\cos^2y+\sin^2 x)}{\sin^2y(1-\cos^2 z-\sin^2 x+\sin^2 x\cos^2 z+\sin^2 x)}$$
$$=\frac{\sin^2z((1-\cos^2 y)+\sin^2 x\cos^2y)}{\sin^2y((1-\cos^2 z)+\sin^2 x\cos^2 z)}$$
$$=\frac{\sin^2z(\sin^2 y+\sin^2 x\cos^2y)}{\sin^2y(\sin^2 z+\sin^2 x\cos^2 z)}$$
$$=\frac{\sin^2z\sin^2 y\left(1+\frac{\sin^2 x\cos^2y}{\sin^2 y}\right)}{\sin^2y\sin^2 z\left(1+\frac{\sin^2 x\cos^2 z}{\sin^2 z}\right)}$$
$$=\frac{1+\frac{\sin^2 x}{\tan^2 y}}{1+\frac{\sin^2 x}{\tan^2 z}}$$
$$=\frac{1+\left(\frac{\sin x}{\tan y}\right)^2}{1+\left(\frac{\sin x}{\tan z}\right)^2}=RHS$$
