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

Hint: 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 $\tan = \dfrac{1}{{\cot }}$