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# $x=p\sec \theta$ and $y=q\tan \theta$ thenA.${{x}^{2}}-{{y}^{2}}={{p}^{2}}{{q}^{2}}$ B.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=pq$ C.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}$D.${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=\dfrac{1}{{{p}^{2}}{{q}^{2}}}$  Verified
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Hint: Out of the given options the right-hand side of the equation is always independent of $\theta$. So, we have to perform some calculations to make left-hand side equations independent of $\theta$ too. Thus, we need to make equations based on $\theta$ and solve it accordingly to remove $\theta$.

Here, we have $x=p\sec \theta$ and $y=q\tan \theta$
Considering each option’s nature, we can see that the right-hand side of the equations are independent of $\theta$. So, we have to form equations based on $\theta$ to eliminate it from these equations.
Now, from the given equations, for $x=p\sec \theta$, we get
$\Rightarrow x=p\sec \theta$
On cross-multiplication, we get
$\Rightarrow \sec \theta =\dfrac{x}{p}...\text{ }\left( 1 \right)$
Similarly, from $y=q\tan \theta$, we have
\begin{align} & \Rightarrow y=q\tan \theta \\ & \Rightarrow \tan \theta =\dfrac{y}{q}...\text{ }\left( 2 \right) \\ \end{align}
Since, we need to make the equations independent of $\theta$, we can use trigonometric identities, i.e.,
${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta$
And, substituting values of $\tan \theta$ and $\sec \theta$ from equation (1) and (2), we get
\begin{align} & \Rightarrow {{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \\ & \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}=1+{{\left( \dfrac{y}{q} \right)}^{2}} \\ \end{align}
By transposing the variables, we get
\begin{align} & \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}=1+{{\left( \dfrac{y}{q} \right)}^{2}} \\ & \Rightarrow {{\left( \dfrac{x}{p} \right)}^{2}}-{{\left( \dfrac{y}{q} \right)}^{2}}=1 \\ & \Rightarrow \dfrac{{{x}^{2}}}{{{p}^{2}}}-\dfrac{{{y}^{2}}}{{{q}^{2}}}=1 \\ \end{align}
Taking LCM on LHS of the equation, we get
\begin{align} & \Rightarrow \dfrac{{{x}^{2}}}{{{p}^{2}}}-\dfrac{{{y}^{2}}}{{{q}^{2}}}=1 \\ & \Rightarrow \dfrac{{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}}{{{p}^{2}}{{q}^{2}}}=1 \\ \end{align}
On cross-multiplication, we get
\begin{align} & \Rightarrow \dfrac{{{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}}{{{p}^{2}}{{q}^{2}}}=1 \\ & \Rightarrow {{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}=1\left( {{p}^{2}}{{q}^{2}} \right) \\ & \Rightarrow {{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}} \\ \end{align}
Hence, on solving the given equations, we get ${{x}^{2}}{{q}^{2}}-{{y}^{2}}{{p}^{2}}={{p}^{2}}{{q}^{2}}$, i.e., option [C] is correct.

Note: The easiest way to solve this kind of problem is by substituting the values of $x$ and $y$ on LHS of every option available and checking it with the values of RHS of that particular option. This method is very useful if the given equations of $x$ and $y$ are very complex.