Prove the following trigonometric identities- cos2-sin-cosec-sin-0

We know that cos^2θ+ sin^2θ = 1 cosec θ = 1/sin θ LHS = cos^2θ/sin θ cosec θ+sin θ = cos^2θ/sin θ 1/sin θ +sin θ = cos^2θ/sin θ 1+sin^2θ/sin θ = cos^2θ 1+sin^ ...

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Standard X

Mathematics

Trigonometric Identities

Prove the fol...

Question

Prove the following trigonometric identities:

$\frac{c o s^{2} θ}{s i n θ} - c o s e c θ + s i n θ = 0$

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Solution

We know that $c o s^{2} θ + s i n^{2} θ = 1$

$c o s e c θ = \frac{1}{s i n θ}$

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

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

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

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

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

$= \frac{1 - 1}{s i n θ}$

$= 0 = R H S$

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$\frac{1 - s i n θ}{1 + s i n θ} = (s e c θ - t a n θ)^{2}$

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Prove the following trigonometric identities: cos2θsin θ−cosec θ+sin θ=0

We know that cos2θ+sin2θ=1 cosec θ=1sin θ LHS=cos2θsin θ−cosec θ+sin θ =cos2θsin θ−1sin θ+sin θ =cos2θsin θ−1+sin2θsin θ =cos2θ−1+sin2θsin θ =cos2θ+sin2θ−1sin θ =1−1sin θ =0=RHS

Prove the following trigonometric identities:

$\frac{c o s^{2} θ}{s i n θ} - c o s e c θ + s i n θ = 0$