If - a sin2 -b sin - cos -c cos2 -1-2a-c -1-2k- then k2 is equal to

A sin^2 θ+b sin θ cos θ+c cos^2 θ 1/2(a+c) =1/2[ acos 2θ + b sin 2θ + c cos 2 θ] =1/2[b sin 2 θ (a c)cos 2 θ] ∵ |b sin 2 θ (a c)cos 2 θ|≤√(b^2+(a c)^2) ∴ | ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

If | a sin2 θ...

Question

$I f ∣ ∣ a s i n^{2} θ + b s i n θ c o s θ + c c o s^{2} θ - \frac{1}{2} (a + c) ∣ ∣ \leq \frac{1}{2} k, t h e n k^{2} i s e q u a l t o$

b^{2} + (a - c)^{2}

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a^{2} + (b - c)^{2}

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c^{2} + (a - b)^{2}

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N o n e o f t h e s e

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Solution

The correct option is A $b^{2} + (a - c)^{2}$
$a s i n^{2} θ + b s i n θ c o s θ + c c o s^{2} θ - \frac{1}{2} (a + c) = \frac{1}{2} [- a c o s 2 θ + b s i n 2 θ + c c o s 2 θ] = \frac{1}{2} [b s i n 2 θ - (a - c) c o s 2 θ] ∵ | b s i n 2 θ - (a - c) c o s 2 θ | \leq \sqrt{b^{2} + (a - c)^{2}} ∴ ∣ ∣ \frac{1}{2} {b s i n 2 θ - (a - c) c o s 2 θ} ∣ ∣ \leq \frac{1}{2} \sqrt{b^{2} + (a - c)^{2}} \Rightarrow ∣ ∣ a s i n^{2} θ + b s i n θ c o s θ + c c o s^{2} θ - \frac{1}{2} (a + c) ∣ ∣ \leq \frac{1}{2} \sqrt{b^{2} + (a - c)^{2}}$

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If∣∣a sin2 θ+b sin θ cos θ+c cos2 θ−12(a+c)∣∣≤12k, then k2 is equal to

$I f ∣ ∣ a s i n^{2} θ + b s i n θ c o s θ + c c o s^{2} θ - \frac{1}{2} (a + c) ∣ ∣ \leq \frac{1}{2} k, t h e n k^{2} i s e q u a l t o$