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The least value of $\cos^{2} θ - 6 \sin θ \cos θ + 3 \sin^{2} θ + 2,$ is

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4 + \sqrt{10}

4 - \sqrt{10}

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detailed solution

Correct option is B

We have,

$\begin{array}{l} y = \cos^{2} θ - 6 \sin θ \cos θ + 3 \sin^{2} θ + 2 \\ \Rightarrow y = (\frac{1 + \cos 2 θ}{2}) - 3 \sin 2 θ + 3 (\frac{1 - \cos 2 θ}{2}) + 2 \\ \Rightarrow y = - \cos 2 θ - 3 \sin 2 θ + 4 \end{array}$

Now,

$- \sqrt{10} \leq - \cos 2 θ - 3 \sin 2 θ \leq \sqrt{10}$ for all $θ$

$\Rightarrow 4 - \sqrt{10} \leq y \leq 4 + \sqrt{10}$

Hence, the least value of $y$ is $4 - \sqrt{10}$ .

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The least value of cos2⁡θ−6sin⁡θcos⁡θ+3sin2⁡θ+2, is