Prove the following trignometric identities- cos-1-sin-1-sin-cos-

We know , cos^2θ + sin^2θ=1 So, Multiplying both numerator and denominator by (1 sinθ), we have cos θ/1+sin θ = cos θ(1 sin θ)/(1+sin θ)(1 sin θ) =cos θ(1 sin ...

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Standard X

Mathematics

Trigonometric Identities

Prove the fol...

Question

Prove the following trignometric identities:

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

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Solution

We know ,

$c o s^{2} θ + s i n^{2} θ = 1$

So, Multiplying both numerator and denominator by (1- sinθ), we have

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

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

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

$= \frac{1 - s i n θ}{c o s θ}$

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Prove the following trignometric identities: cos θ1+sin θ=1−sin θcos θ

We know , cos2θ+sin2θ=1 So, Multiplying both numerator and denominator by (1- sinθ), we have cos θ1+sin θ=cos θ(1−sin θ)(1+sin θ)(1−sin θ) =cos θ(1−sin θ)(1−sin2θ) = cos θ(1−sin θ)cos2θ =1−sin θcos θ

Prove the following trignometric identities:

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