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

We know , sin2θ+cos2θ=1 Multiplying both numerator and denominator by (1+cosθ), we have =1−cos2θ/(1+cosθ)(sinθ) =(sin2θ)/(1+cosθ)(sinθ) =(sinθ)/(1+cosθ) ...

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Standard X

Mathematics

Relation between Trigonometric Ratios

Prove the fol...

Question

Prove the following trigonometric identities:

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

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Solution

We know ,

sin2θ+cos2θ=1

Multiplying both numerator and denominator by (1+cosθ), we have

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

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Prove the following trigonometric identities: 1−cos θsin θ=sin θ1+cos θ

We know , sin2θ+cos2θ=1 Multiplying both numerator and denominator by (1+cosθ), we have =1−cos2θ(1+cosθ)(sinθ)=(sin2θ)(1+cosθ)(sinθ)=(sinθ)(1+cosθ)

Prove the following trigonometric identities:

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

We know ,

sin2θ+cos2θ=1

Multiplying both numerator and denominator by (1+cosθ), we have

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