Question

If we are given the equations \[\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = 1\] and \[\dfrac{x}{a}\sin \theta - \dfrac{y}{b}\cos \theta = 1\], prove that \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 2\]

Answer 1

Hint: In this question, we have given two-equations from that equation we have to prove another relation. So first we square both the equations on both sides and then add both the equations and on further simplify use the identity of trigonometry then we get the same relation which we want to prove.

Complete step-by-step solution:
Given,
Two equation are given
\[\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = 1\] and
\[\dfrac{x}{a}\sin \theta - \dfrac{y}{b}\cos \theta = 1\]
To prove,
\[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 2\]
To find this relation we do following steps
\[\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta = 1\] ………..…….(i)
On squaring both side
\[{\left( {\dfrac{x}{a}\cos \theta + \dfrac{y}{b}\sin \theta } \right)^2} = {1^2}\]
Using the identity \[{\left( {a + b} \right)^2}\]
\[\dfrac{{{x^2}}}{{{a^2}}}{\cos ^2}\theta + \dfrac{{{y^2}}}{{{b^2}}}{\sin ^2}\theta + 2\dfrac{x}{a}\dfrac{y}{b}\sin \theta \cos \theta = 1\] …………………(ii)
Now taking next equation that is given in question
\[\dfrac{x}{a}\sin \theta - \dfrac{y}{b}\cos \theta = 1\] …………………(iii)
On squaring both side
\[{\left( {\dfrac{x}{a}\sin \theta - \dfrac{y}{b}\cos \theta } \right)^2} = {1^2}\]
Using the identity \[{\left( {a - b} \right)^2}\]
\[\dfrac{{{x^2}}}{{{a^2}}}{\sin ^2}\theta + \dfrac{{{y^2}}}{{{b^2}}}{\cos ^2}\theta -2\dfrac{x}{a}\dfrac{y}{b}\sin \theta \cos \theta = 1\] ……………(iv)
Now adding equation (ii) and (iv)
\[\dfrac{{{x^2}}}{{{a^2}}}{\cos ^2}\theta + \dfrac{{{y^2}}}{{{b^2}}}{\sin ^2}\theta + 2\dfrac{x}{a}\dfrac{y}{b}\sin \theta \cos \theta + \dfrac{{{x^2}}}{{{a^2}}}{\sin ^2}\theta + \dfrac{{{y^2}}}{{{b^2}}}{\cos ^2}\theta - 2\dfrac{x}{a}\dfrac{y}{b}\sin \theta \cos \theta = 1 + 1\]
On further solving
\[\dfrac{{{x^2}}}{{{a^2}}}{\cos ^2}\theta + \dfrac{{{y^2}}}{{{b^2}}}{\sin ^2}\theta + \dfrac{{{x^2}}}{{{a^2}}}{\sin ^2}\theta + \dfrac{{{y^2}}}{{{b^2}}}{\cos ^2}\theta = 2\]
Taking some terms common
\[\dfrac{{{x^2}}}{{{a^2}}}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + \dfrac{{{y^2}}}{{{b^2}}}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 2\]
Now using the identity of trigonometry \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
On putting this value
\[\dfrac{{{x^2}}}{{{a^2}}}\left( 1 \right) + \dfrac{{{y^2}}}{{{b^2}}}\left( 1 \right) = 2\]
On further solving
\[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 2\]
Hence proved.
Additional information:
Identity used in order to solve this question:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\],
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]

Note: To solve this type of question we have to eliminate trigonometric function. So for eliminating try to use trigonometry identity that are
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\],
\[1 + {\tan ^2}\theta = {\sec ^2}\theta \] and
\[1 + {\cot ^2}\theta = co{\sec ^2}\theta \]
We have to square both sides. If we are not able to use this identity directly then make some changes. You may commit mistakes like directly adding both equations and trying to eliminate trigonometric functions from there. Use the correct identities to make calculations simpler and easier.