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Question: Eliminate \[\theta \], \[\begin{aligned} & x=2\sec \theta +3\tan \theta \\\ & y=3\sec \th...

Eliminate θ\theta ,

& x=2\sec \theta +3\tan \theta \\\ & y=3\sec \theta -2\tan \theta \\\ \end{aligned}$$
Explanation

Solution

Hint : Square the two equations and add the resultant equation to eliminate the θ\theta .
Formulas used:
The standard value of sec2θ+tan2θ=1{{\sec }^{2}}\theta +{{\tan }^{2}}\theta =1

Complete step-by-step answer :
First step will be squaring both the equation,
The next step is to add the following equation and the solution can be obtained.

The θ\theta can be eliminating as,

& x=2\sec \theta +3\tan \theta \text{ }\left( 1 \right) \\\ & y=3\sec \theta -2\tan \theta \text{ }\left( 2 \right) \\\ & \text{Solving equation }\left( 1 \right)\text{ and }\left( 2 \right), \\\ & \text{Squaring both side of equation }\left( 1 \right), \\\ & {{x}^{2}}={{\left( 2\sec \theta +3\tan \theta \right)}^{2}} \\\ & {{x}^{2}}\text{=4}{{\sec }^{2}}\theta +9{{\tan }^{2}}\theta +12\sec \theta \tan \theta \text{ }\left( 3 \right) \\\ & \text{Squaring both side of equation }\left( 2 \right), \\\ & {{y}^{2}}={{\left( 3\sec \theta -2\tan \theta \right)}^{2}} \\\ & {{y}^{2}}\text{=4}{{\sec }^{2}}\theta +9{{\tan }^{2}}\theta -12\sec \theta \tan \theta \text{ }\left( 4 \right) \\\ & \text{Adding equation }\left( 3 \right)\text{ and }\left( 4 \right), \\\ & {{x}^{2}}\text{+}{{y}^{2}}\text{=}\left( \text{4}{{\sec }^{2}}\theta +9{{\tan }^{2}}\theta +12\sec \theta \tan \theta \right)+\left( \text{4}{{\sec }^{2}}\theta +9{{\tan }^{2}}\theta -12\sec \theta \tan \theta \right) \\\ & {{x}^{2}}\text{+}{{y}^{2}}=\text{4}{{\sec }^{2}}\theta +9{{\tan }^{2}}\theta +12\sec \theta \tan \theta +\text{4}{{\sec }^{2}}\theta +9{{\tan }^{2}}\theta -12\sec \theta \tan \theta \\\ & {{x}^{2}}\text{+}{{y}^{2}}=\text{4}\left( {{\sec }^{2}}\theta +{{\tan }^{2}}\theta \right)+9\left( {{\sec }^{2}}\theta +{{\tan }^{2}}\theta \right) \\\ & {{x}^{2}}\text{+}{{y}^{2}}=4\cdot 1+9\cdot 1 \\\ & {{x}^{2}}\text{+}{{y}^{2}}=13 \\\ \end{aligned}$$ Thus, the required solution is $${{x}^{2}}\text{+}{{y}^{2}}=13$$. **Note** : In the elimination method we have to either add or subtract the given equations to get an equation in one variable, when the coefficients of one variable are opposites, we have to square the equation and add it to another equation to eliminate the variable.