Sin-2 sin 3-2 cos 3-cos-

The correct option is D tan θ sin θ 2 sin^3 θ/2 cos^3 θ cos θ =sin θ(1 2 sin^2 θ)/cos θ(2 cos^2 θ 1) =tan θ(sin^2 θ+cos^2 θ 2 sin^2 θ)/(2 cos^2 θ sin^2 θ ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

sinθ-2 sin 3θ...

Question

$\frac{s i n θ - 2 s i n^{3} θ}{2 c o s^{3} θ - c o s θ} =$

$c o s e c θ$

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$c o s θ$

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$5 t a n θ$

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$t a n θ$

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Solution

The correct option is D
$t a n θ$

$\frac{s i n θ - 2 s i n^{3} θ}{2 c o s^{3} θ - c o s θ}$

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

$= \frac{t a n θ (s i n^{2} θ + c o s^{2} θ - 2 s i n^{2} θ)}{(2 c o s^{2} θ - s i n^{2} θ - c o s^{2} θ)}$

$= \frac{t a n θ (c o s^{2} θ - s i n^{2} θ)}{(c o s^{2} θ - s i n^{2} θ)}$

$= t a n θ$

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Similar questions

Q. Solve:

\frac{sin θ - 2 {sin}^{3} θ}{2 {cos}^{3} θ - cos θ}

Show that:

\frac{sin θ - 2 {sin}^{3} θ}{2 {cos}^{3} θ - cos θ} = tan θ

Q. Prove that

\frac{sin θ - 2 {sin}^{3} θ}{2 {cos}^{3} θ - cos θ} = tan θ

p r o v e t h a t \frac{s i n θ - 2 s {i n}^{3} θ}{2 c o s^{3} θ - c o s θ} = t a n θ

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sin θ−2 sin3 θ2 cos3 θ−cos θ=

The correct option is D tan θ sin θ−2 sin3 θ2 cos3 θ−cos θ =sin θ(1−2 sin2 θ)cos θ(2 cos2 θ−1) =tan θ(sin2 θ+cos2 θ−2 sin2 θ)(2 cos2 θ−sin2 θ−cos2 θ) =tan θ(cos2 θ−sin2 θ)(cos2 θ−sin2 θ) =tan θ

$\frac{s i n θ - 2 s i n^{3} θ}{2 c o s^{3} θ - c o s θ} =$