Question

Eliminate $t$ in $p\tan t+q\sec t-x=0,{{p}^{'}}\tan t+{{q}^{'}}\sec t-y=0$ where $p,{{p}^{'}},q,{{q}^{'}},x,y$ are real numbers.

Answer 1

Hint: We multiply ${{q}^{'}}$ with first equation and $q$ with second equation in order to eliminate $\sec t$. We find $\tan t$ in terms of the given constant terms here and which we put in first equation to get $\sec t$. We use the trigonometric formula ${{\sec }^{2}}t-{{\tan }^{2}}t=1$ and put $\tan t,\sec t$to eliminate $t$.

Complete step-by-step solution:
We know that in the method of elimination when we are asked to eliminate a particular variable of constant we have to find a relation among the rest of the variables and constants.
The given pair of equations are
\[\begin{align}
  & p\tan t+q\sec t-x=0...(1) \\
 & {{p}^{'}}\tan t+{{q}^{'}}\sec t-y=0...(2) \\
\end{align}\]
We are asked to eliminate $t$ here which is present in $\tan t$ and $\sec t.$ So we need to find a relation among rest of the real constants $p,{{p}^{'}},q,{{q}^{'}},x,y$ free of $\tan t$ and $\sec t.$ We first multiply ${{q}^{'}}$ with equation (1) and $q$ with equation (2) in order to eliminate $\sec t$. We have
\[\begin{align}
  & p{{q}^{'}}\tan t+q{{q}^{'}}\sec t-x{{q}^{'}}=0...(1) \\
 & {{p}^{'}}q\tan t+q{{q}^{'}}\sec t-yq=0...(2) \\
\end{align}\]
We now subtract equation(2) from equation (1) and get
\[\begin{align}
  & \left( p{{q}^{'}}-{{p}^{'}}q \right)\tan t-\left( x{{q}^{'}}-yq \right)=0 \\
 & \Rightarrow \tan t=\dfrac{x{{q}^{'}}-yq}{p{{q}^{'}}-{{p}^{'}}q} \\
\end{align}\]
We put the value of $\tan t$ in equation (1) and get ,
\[\begin{align}
  & p\left( \dfrac{x{{q}^{'}}-yq}{p{{q}^{'}}-{{p}^{'}}q} \right)+q\sec t-x=0 \\
 & \Rightarrow q\sec t=x-p\left( \dfrac{x{{q}^{'}}-yq}{p{{q}^{'}}-{{p}^{'}}q} \right) \\
 & \Rightarrow q\sec t=\dfrac{xp{{q}^{'}}-x{{p}^{'}}q-xp{{q}^{'}}+ypq}{p{{q}^{'}}-{{p}^{'}}q} \\
 & \Rightarrow q\sec t=\dfrac{ypq-x{{p}^{'}}q}{p{{q}^{'}}-{{p}^{'}}q} \\
\end{align}\]
We divide both side by $q$ and get ,
\[\sec t=\dfrac{yp-x{{p}^{'}}}{p{{q}^{'}}-{{p}^{'}}q}\]
We know from the trigonometric identity that
\[{{\sec }^{2}}t-{{\tan }^{2}}t=1\]
We put the values of $\sec t$ and $\tan t$ and proceed ,
\[{{\left( \dfrac{yp-x{{p}^{'}}}{p{{q}^{'}}-{{p}^{'}}q} \right)}^{2}}-{{\left( \dfrac{x{{q}^{'}}-yq}{p{{q}^{'}}-{{p}^{'}}q} \right)}^{2}}=1\]
We multiply ${{\left( p{{q}^{'}}-{{p}^{'}}q \right)}^{2}}$ in all of the above terms and get ,
\[{{\left( yp-x{{p}^{'}} \right)}^{2}}-{{\left( x{{q}^{'}}-yq \right)}^{2}}={{\left( p{{q}^{'}}-{{p}^{'}}q \right)}^{2}}\]
The above obtained equation is free of $\sec t,\tan t$ and hence is also free of $t$. So it is the required equation.

Note: We take note of that the given real constants $p,{{p}^{'}},q,{{q}^{'}}$ are non-zero as well the $p{{q}^{'}}-{{p}^{'}}q$ is nonzero. We can also begin with eliminating $\tan t$ and then find $\sec t$. We know that when are given a pair of linear equations ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0,{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0$ we can directly find unknowns as $x=\dfrac{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}},y=\dfrac{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}$ where ${{a}_{1}}{{b}_{2}}-{{a}_{1}}{{b}_{2}}$ is non-zero.