2 sin-cos-cos-1-sin-sin 2-cos 2-

LHS 2 sin θ cos θ cos θ/1 sin θ +sin^2 θ cos^2 θ=cos θ(2 sin θ 1)/1 cos^2 θ +sin^2 θ sin θ =cos θ(2 sin θ 1)/sin^2 θ +sin^2 θ sin θ[∵ 1 cos^2 θ=sin^2 θ] =cos ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

2 sinθcosθ-co...

Question

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

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Solution

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

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

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

Hence proved.

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Similar questions

If cos $θ = \frac{5}{13}$ , find the value of $\frac{s i n^{2} θ - c o s^{2} θ}{2 s i n θ c o s θ} \times \frac{1}{t a n^{2} θ}$

Q. Prove that

\frac{sin θ - cos θ}{sin θ + cos θ} + \frac{sin θ + cos θ}{sin θ - cos θ} = \frac{2}{{sin}^{2} θ - {cos}^{2} θ}

Q. Solve the following equations:
(i)

\cos θ + \cos 2 θ + \cos 3 θ = 0

(ii)

\cos θ + \cos 3 θ - \cos 2 θ = 0

(iii)

\sin θ + \sin 5 θ = \sin 3 θ

(iv)

\cos θ \cos 2 θ \cos 3 θ = \frac{1}{4}

(v)

\cos θ + \sin θ = \cos 2 θ + \sin 2 θ

(vi)

\sin θ + \sin 2 θ + \sin 3 = 0

(vii)

\sin θ + \sin 2 θ + \sin 3 θ + \sin 4 θ = 0

(viii)

\sin 3 θ - \sin θ = 4 \cos^{2} θ - 2

(ix)

\sin 2 θ - \sin 4 θ + \sin 6 θ = 0

\frac{s i n θ + s i n 2 θ}{1 + c o s θ + c o s 2 θ}

is equal to

Q. Solve:

\frac{sin θ}{sin 2 θ} = \frac{cos θ}{cos 2 θ}

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2sinθcosθ−cosθ1−sinθ+sin2θ−cos2θ=cotθ

LHS2sinθcosθ−cosθ1−sinθ+sin2θ−cos2θ=cosθ(2sinθ−1)1−cos2θ+sin2θ−sinθ =cosθ(2sinθ−1)sin2θ+sin2θ−sinθ[∵1−cos2θ=sin2θ] =cosθ(2sinθ−1)2sin2θ−sinθ=cosθ(2sinθ−1)sin(2sinθ−1)=cosθsinθ=cotθ=RHS Hence proved.

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