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The maximum value of the expression $\dfrac{1}{{{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta }$ isA. 0B. 1C. 2D. 3E. 4

Last updated date: 19th Jun 2024
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Hint: We need to find the minimum value of ${{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta$ to find the maximum value of the expression $\dfrac{1}{{{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta }$ . First, simplify the expression using trigonometric formulas. Then we try to convert the equation into an equation of singular ratio using the formula of $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ . Then we apply the condition of minimum value to get the answer.

Complete step by step answer:
First, we try to find the simplified form of the equation ${{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta$ .
We have the identity theorem of ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ .
We convert the equation as
$\Rightarrow {{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta$
$=\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)+\dfrac{3}{2}\left( 2\sin \theta \cos \theta \right)+4{{\cos }^{2}}\theta$
$=1+\dfrac{3}{2}\left( 2\sin \theta \cos \theta \right)+2\left( 2{{\cos }^{2}}\theta \right)$
Now we apply the theorems of $\sin 2\theta =2\sin \theta \cos \theta$ and $1+\cos 2\theta =2{{\cos }^{2}}\theta$.
$\Rightarrow 1+\dfrac{3}{2}\left( 2\sin \theta \cos \theta \right)+2\left( 2{{\cos }^{2}}\theta \right)$
$=1+\dfrac{3}{2}\left( \sin 2\theta \right)+2\left( 1+\cos 2\theta \right)$
$=3+\dfrac{3\sin 2\theta }{2}+2\cos 2\theta$
To find the maximum value of the expression $\dfrac{1}{{{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta }$ , we need to find the minimum value of ${{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta$ which is equal to $3+\dfrac{3\sin 2\theta }{2}+2\cos 2\theta$ .
Now we try to convert the equation into one particular ratio.
We will take $\dfrac{5}{2}$ common from sin and cos terms and get
$3+\dfrac{3\sin 2\theta }{2}+2\cos 2\theta =3+\dfrac{5}{2}\left( \dfrac{3}{5}\sin 2\theta +\dfrac{4}{5}\cos 2\theta \right)$ .
We assume $\cos \alpha =\dfrac{3}{5}$ which gives $\sin \alpha =\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}=\sqrt{1-\dfrac{9}{25}}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}$ .
The equation becomes $3+\dfrac{5}{2}\left( \dfrac{3}{5}\sin 2\theta +\dfrac{4}{5}\cos 2\theta \right)=3+\dfrac{5}{2}\left( \sin 2\theta \cos \alpha +\cos 2\theta \sin \alpha \right)$ .
Now we apply $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$ . We assume $A=2\theta ;B=\alpha$ .
$3+\dfrac{5}{2}\left( \sin 2\theta \cos \alpha +\cos 2\theta \sin \alpha \right)=3+\dfrac{5}{2}\sin \left( 2\theta +\alpha \right)$.
Now we know that for any value of $x\in \mathbb{R}$, the minimum value of $\sin x$ is -1.
This means the minimum value of $\sin \left( 2\theta +\alpha \right)$ is $-1$. We will find the minimum value of $3+\dfrac{5}{2}\sin \left( 2\theta +\alpha \right)$ by putting minimum value of $\sin \left( 2\theta +\alpha \right)$ as $-1$.
So,$3+\dfrac{5}{2}\sin \left( 2\theta +\alpha \right)=$$3+\dfrac{5}{2}\times(-1)$
$=\dfrac{1}{2}$
Therefore, the minimum value of $3+\dfrac{3\sin 2\theta }{2}+2\cos 2\theta$ which is equal to ${{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta$ , is $\dfrac{1}{2}$ .
The maximum value of the expression $\dfrac{1}{{{\sin }^{2}}\theta +3\sin \theta \cos \theta +5{{\cos }^{2}}\theta }$ is $\dfrac{1}{\dfrac{1}{2}}=2$ .
The correct option is (C).

Note:
The general formula of finding minimum and maximum value of equation types like $a\sin \alpha +b\cos \alpha$ is $-\sqrt{{{a}^{2}}+{{b}^{2}}}\le \left( a\sin \alpha +b\cos \alpha \right)\le \sqrt{{{a}^{2}}+{{b}^{2}}}$ . For any values of $x\in \mathbb{R}$ , minimum and maximum values of ratios like $\sin x,\cos x$ is $-1\le \sin x,\cos x\le 1$ .