Prove that- sin -2 sin3 -2 cos3 -cos -tan -4 Marks-

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) [0.5 Marks] =tan θ(sin^2 θ+cos^2 θ 2 sin^2 θ)/(2 cos^2 θ sin^2 θ cos^2 θ) [2 Marks] = ...

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Standard XII

Mathematics

Iota

Prove that: s...

Question

Prove that: $\frac{s i n θ - 2 s i n^{3} θ}{2 c o s^{3} θ - c o s θ} = t a n θ$
[4 Marks]

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Solution

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

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

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

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

$= t a n θ$
[0.5 Marks]

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Q. Prove that

\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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Q. Prove that

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Prove that: sin θ−2 sin3 θ2 cos3 θ−cos θ=tan θ [4 Marks]

sin θ−2 sin3 θ2 cos3 θ−cos θ =sin θ(1−2 sin2 θ)cos θ(2 cos2 θ−1) [0.5 Marks] =tan θ(sin2 θ+cos2 θ−2 sin2 θ)(2 cos2 θ−sin2 θ−cos2 θ) [2 Marks] =tan θ(cos2 θ−sin2 θ)(cos2 θ−sin2 θ) [1 Mark] =tan θ [0.5 Marks]

Prove that: $\frac{s i n θ - 2 s i n^{3} θ}{2 c o s^{3} θ - c o s θ} = t a n θ$
[4 Marks]