Question

If $2y\cos \theta = x\sin \theta$ and $2x\sec \theta - y\cos ec\theta = 3$ then ${x^2} + 4{y^2} =$
1. $4$
2. $- 4$
3. $\pm 4$
4. None of these

Answer 1

First, we need to analyze the given information which is in the trigonometric form.

The trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
From the given we asked to calculate the value ${x^2} + 4{y^2} =$?, so we need to know the formulas in sine, cos, sec, cosec in the trigonometry.
Formula used:
${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\sec \theta = \dfrac{1}{{\cos \theta }},\cos ec\theta = \dfrac{1}{{\sin \theta }}$

Complete step-by-step solution:
Since from the given that we have to trigonometric functions $2y\cos \theta = x\sin \theta$ and $2x\sec \theta - y\cos ec\theta = 3$
Let us the function $2y\cos \theta = x\sin \theta$ as an equation $(1)$
Now take the second function, which is given as $2x\sec \theta - y\cos ec\theta = 3$ and since we know that $\sec \theta = \dfrac{1}{{\cos \theta }},\cos ec\theta = \dfrac{1}{{\sin \theta }}$, so apply these values in the given function then we get $2x\sec \theta - y\cos ec\theta = 3 \Rightarrow 2x\dfrac{1}{{\cos \theta }} - y\dfrac{1}{{\sin \theta }} = 3$
Now cross multiplying these values, then we get $2x\dfrac{1}{{\cos \theta }} - y\dfrac{1}{{\sin \theta }} = 3 \Rightarrow \dfrac{{2x\sin \theta - y\cos \theta }}{{\sin \theta \cos \theta }} = 3$
Further solving the function, we get $2x\sin \theta - y\cos \theta = 3\sin \theta \cos \theta$
Now substitute the equation $(1)$ in the above equation then we get; $2x\sin \theta - y\cos \theta = 3\sin \theta \cos \theta \Rightarrow 2(2y\cos \theta ) - y\cos \theta = 3\sin \theta \cos \theta$
Further solving, $4y\cos \theta - y\cos \theta = 3\sin \theta \cos \theta \Rightarrow 3y\cos \theta = 3\sin \theta \cos \theta$ and canceling the common terms and thus we have $3y\cos \theta = 3\sin \theta \cos \theta \Rightarrow y = \sin \theta$
Now substitute the value of $y = \sin \theta$ in the equation $(1)$ then we get $2y\cos \theta = x\sin \theta \Rightarrow 2\sin \theta \cos \theta = x\sin \theta$ again canceling the common terms, we get $2\sin \theta \cos \theta = x\sin \theta \Rightarrow \cos \theta = \dfrac{x}{2}$
Hence, we have the values $y = \sin \theta$ and $\cos \theta = \dfrac{x}{2}$. Now apply these values in the general equation ${\sin ^2}\theta + {\cos ^2}\theta = 1$ then we get ${\sin ^2}\theta + {\cos ^2}\theta = 1 \Rightarrow {y^2} + {(\dfrac{x}{2})^2} = 1$
Further solving we get ${y^2} + \dfrac{{{x^2}}}{{{2^2}}} = 1 \Rightarrow \dfrac{{4{y^2} + {x^2}}}{4} = 1 \Rightarrow 4{y^2} + {x^2} = 4$ which is the required value.
Hence option $A)4$ is correct.

Note: Simply using the trigonometric value of sine and cos for the sec and cosec we solved the given function.
Where sine and cos can also be written as $\sec \theta = \dfrac{1}{{\cos \theta }},\cos ec\theta = \dfrac{1}{{\sin \theta }} \Rightarrow \cos \theta = \dfrac{1}{{\sec \theta }},\sin \theta = \dfrac{1}{{\cos ec\theta }}$ because they are interrelated.
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan$ and $\tan = \dfrac{1}{{\cot }}$