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

LHS=1+cosθ sin^2θsinθ (1+cosθ) #160; #160; #160; =(1+cosθ) (1 cos^2θ)sinθ(1+cosθ) =(1+cosθ) ((1 cosθ)(1+cosθ))sinθ(1+cosθ) =(1+cosθ ...

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General Solution of Trigonometric Equation

Solve 1+cosθ...

Question

Solve
$\frac{1 + cos θ - {sin}^{2} θ}{sin θ (1 + cos θ)} = cot θ$

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Solution

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Prove that:

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

Prove the following trigonometric identities:

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

\cot θ = \frac{15}{8} then evaluate \frac{(1 + \sin θ) (1 - \sin θ)}{(1 + \cos θ) (1 - \cos θ)}

\frac{1}{sec θ - tan θ} - \frac{1}{cos θ} = \frac{1}{cos θ} - \frac{1}{sec θ + tan θ}

Prove the following trigonometric identities.
(i) $\frac{1 + cos θ + sin θ}{1 + cos θ - sin θ} = \frac{1 + sin θ}{cos θ}$

(ii) $\frac{sin θ - cos θ + 1}{sin θ + cos θ - 1} = \frac{1}{sec θ - tan θ}$

(iii) $\frac{cos θ - sin θ + 1}{cos θ + sin θ - 1} = cosec θ + cot θ$

(iv) $(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ$

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