Sin 3 -1-2 cos 2 - is equal to

The correct option is C sin θ We have, sin θ/1+2 cos 2θ=3 sin θ 4 sin^2 θ/1+2 (1 2 sin^2 θ) = 3 sin θ 4 sin^2 θ/1+2 4 sin^2 θ = sin θ (3 4 sin^2 θ)/(3 ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

sin 3 θ/1+2 c...

Question

$\frac{s i n 3 θ}{1 + 2 c o s 2 θ}$ is equal to

$c o s θ$

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$s i n θ$

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$- c o s θ$

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$s i n θ$

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Solution

The correct option is B
$s i n θ$

We have,

$\frac{s i n θ}{1 + 2 c o s 2 θ} = \frac{3 s i n θ - 4 s i n^{2} θ}{1 + 2 (1 - 2 s i n^{2} θ)} = \frac{3 s i n θ - 4 s i n^{2} θ}{1 + 2 - 4 s i n^{2} θ} = \frac{s i n θ (3 - 4 s i n^{2} θ)}{(3 - 4 s i n^{2} θ)} = s i n θ$

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sin 3θ1+2 cos 2θ is equal to

The correct option is B sin θ We have, sin θ1+2 cos 2θ=3 sin θ−4 sin2θ1+2(1−2 sin2θ)=3 sin θ−4 sin2θ1+2−4 sin2θ=sin θ(3−4 sin2θ)(3−4 sin2θ)=sin θ

$\frac{s i n 3 θ}{1 + 2 c o s 2 θ}$ is equal to

The correct option is B
$s i n θ$

We have,

$\frac{s i n θ}{1 + 2 c o s 2 θ} = \frac{3 s i n θ - 4 s i n^{2} θ}{1 + 2 (1 - 2 s i n^{2} θ)} = \frac{3 s i n θ - 4 s i n^{2} θ}{1 + 2 - 4 s i n^{2} θ} = \frac{s i n θ (3 - 4 s i n^{2} θ)}{(3 - 4 s i n^{2} θ)} = s i n θ$