Question

If \[x\sec \theta = 1 \pm y\tan \theta \] and \[{x^2}{\sec ^2}\theta = 5 + {y^2}{\tan ^2}\theta \] then the relationship between \['x'\] and \['y'\] is
A. \[{x^2} + {y^2} = 9\]
B. \[{x^2}{y^2} + 4({x^2} - {y^2}) = 0\]
C. \[{x^2}{y^2} - 9{y^2} = 0\]
D. \[{x^2}{y^2} + 4{x^2} - 9{y^2} = 0\]

Answer 1

Hint: To solve this problem we need to find the value of the variables by using the two given equations. Substitute the values of one equation in another equation and use the formulas to simplify the equation to get the required answer.

Formula used: \[{\sec ^2}\theta = (1 + {\tan ^2}\theta )\]
\[\dfrac{1}{{\sec \theta }} = \cos \theta \]
\[{(a - b)^2} = {a^2} - 2ab + {b^2}\]

Complete step-by-step answer:
We first consider the equation \[x\sec \theta = 1 \pm y\tan \theta \]
It can be written as follows when taking the 1 to left hand side.
\[x\sec \theta - 1 = \pm y\tan \theta \]
By squaring on both sides we get
\[{(x\sec \theta - 1)^2} = {( \pm y\tan \theta )^2}\]
Firstly, \[{(x\sec \theta - 1)^2}\] is in the form\[{(a - b)^2} = {a^2} - 2ab + {b^2}\]. Here \[a = x\sec \theta \]and\[b = 1\].
By expanding \[{(x\sec \theta - 1)^2}\] we get \[{x^2}{\sec ^2}\theta - 2x\sec \theta + 1 \to (A)\]
Secondly, \[{( \pm y\tan \theta )^2}\] can be taken as \[{y^2}{\tan ^2}\theta \to ({\rm B})\]
This is because when we square \[ \pm \]we always end up with a positive term.
From the (A) and (B) we have,
\[ \Rightarrow {x^2}{\sec ^2}\theta - 2x\sec \theta + 1 = {y^2}{\tan ^2}\theta \]
Since it is given that \[{x^2}{\sec ^2}\theta = 5 + {y^2}{\tan ^2}\theta \Rightarrow {x^2}{\sec ^2}\theta - 5 = {y^2}{\tan ^2}\theta \] we can use this in the above equation and reduce as,
\[ \Rightarrow {x^2}{\sec ^2}\theta - 2x\sec \theta + 1 = {x^2}{\sec ^2}\theta - 5\]
\[ \Rightarrow {x^2}{\sec ^2}\theta - 2x\sec \theta + 1 - {x^2}{\sec ^2}\theta = - 5\]
Now, \[{x^2}{\sec ^2}\theta \] term will get cancelled as they have opposite signs.
\[ \Rightarrow - 2x\sec \theta + 1 = - 5\]
Taking 1 to right hand side and solving,
\[ \Rightarrow - 2x\sec \theta = - 5 - 1\]
\[ \Rightarrow - 2x\sec \theta = - 6\]
The minus signs on both sides can now be cancelled and we also divide 6 on the right-hand side by the 2 from the left-hand side.
\[ \Rightarrow x\sec \theta = \dfrac{6}{2}\]
Dividing 6 by 2 leaves a quotient of 3 on the right hand side
\[ \Rightarrow x\sec \theta = 3\]
Since we know that\[\dfrac{1}{{\sec \theta }} = \cos \theta \]
\[ \Rightarrow x = 3 \times \dfrac{1}{{\sec \theta }}\]
\[ \Rightarrow x = 3\cos \theta \]
Now we can substitute \[x = 3\cos \theta \] in \[x\sec \theta = 1 \pm y\tan \theta \] to find \[{\tan ^2}\theta \].
\[ \Rightarrow 3\cos \theta \sec \theta = 1 \pm y\tan \theta \]
As we have already seen, \[\dfrac{1}{{\sec \theta }} = \cos \theta \] the above equation can be written as
\[ \Rightarrow 3 \times \dfrac{1}{{\sec \theta }} \times \sec \theta = 1 \pm y\tan \theta \]
Cancelling \[\dfrac{1}{{\sec \theta }} \times \sec \theta \] we get 1 and multiply 1 with 3, we get 3.
\[ \Rightarrow 3 = 1 \pm y\tan \theta \]
Taking 1 to left had side we solve as,
\[ \Rightarrow 3 - 1 = \pm y\tan \theta \]
\[ \Rightarrow 2 = \pm y\tan \theta \]
On squaring \[2 = \pm y\tan \theta \]on both sides we get \[4 = {y^2}{\tan ^2}\theta \]
\[ \Rightarrow 4 = {y^2}{\tan ^2}\theta \]
\[ \Rightarrow \dfrac{4}{{{y^2}}} = {\tan ^2}\theta \to (C)\]
Now consider \[{x^2}{\sec ^2}\theta = 5 + {y^2}{\tan ^2}\theta \] which is given in the question.
\[ \Rightarrow {x^2}(1 + {\tan ^2}\theta ) = 5 + {y^2}{\tan ^2}\theta \]
We know that \[{\sec ^2}\theta = (1 + {\tan ^2}\theta )\]
\[ \Rightarrow {x^2} + {x^2}{\tan ^2}\theta = 5 + {y^2}{\tan ^2}\theta \]
By bringing all the terms to left hand side, we get
\[ \Rightarrow {x^2} + {x^2}{\tan ^2}\theta - 5 - {y^2}{\tan ^2}\theta = 0\]
Since \[{\tan ^2}\theta \] is a common term for both \[{x^2}\] as well as \[{y^2}\] we can further reduce the equation as follows.
\[ \Rightarrow {x^2} + ({x^2} - {y^2}){\tan ^2}\theta - 5 = 0\]
as we already have \[\dfrac{4}{{{y^2}}} = {\tan ^2}\theta \] from equation (C) we can use it in the above equation.
\[ \Rightarrow {x^2} + ({x^2} - {y^2})\dfrac{4}{{{y^2}}} - 5 = 0\]
On multiplying \[\dfrac{4}{{{y^2}}}\] with \[({x^2} - {y^2})\] and simplifying we get,
\[ \Rightarrow {x^2} + \left( {\dfrac{{4{x^2}}}{{{y^2}}} - \dfrac{{4{y^2}}}{{{y^2}}}} \right) - 5 = 0\]
\[ \Rightarrow {x^2} + \dfrac{{4{x^2}}}{{{y^2}}} - 4 - 5 = 0\]
\[ \Rightarrow {x^2} + \dfrac{{4{x^2}}}{{{y^2}}} - 9 = 0\]
Taking \[{y^2}\] as LCM we can find that
\[ \Rightarrow \dfrac{{{y^2}{x^2} + 4{x^2} - 9{y^2}}}{{{y^2}}} = 0\]
\[ \Rightarrow {x^2}{y^2} + 4{x^2} - 9{y^2} = 0\].
Hence the solution.
If \[x\sec \theta = 1 \pm y\tan \theta \] and \[{x^2}{\sec ^2}\theta = 5 + {y^2}{\tan ^2}\theta \] then the relationship between\['x'\] and \['y'\] is
\[{x^2}{y^2} + 4{x^2} - 9{y^2} = 0\].

Option D is the correct answer.

Note: In the question, students make a mistake that has to change the term with \[ \pm \] as its sign, when we have to square the sign, it will always result in a positive term.
Because \[( + ) \times ( + ) = + \]and \[( - ) \times ( - ) = + \]
Some of the formula used in the trigonometric functions,
\[\dfrac{1}{{\operatorname{cosec} \theta }} = \sin \theta \]
\[\dfrac{1}{{{\text{cot}}\theta }} = \tan \theta \]