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Calculating Heights and Distances

Prove that: θ...

Question

Prove that: $\frac{\sec θ + \tan θ - 1}{\tan θ - \sec θ + 1} = \frac{\cos θ}{1 - \sin θ}$

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\frac{secθ + tanθ - 1}{tanθ - secθ + 1} = \frac{cosθ}{1 - sinθ}

\frac{secθ cosec (90 ° - θ) - tanθ \cot (90 ° - θ) + \sin^{2} 55 ° + \sin^{2} 35 °}{\tan 10 ° \tan 20 ° \tan 60 ° \tan 70 ° \tan 80 °}

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

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

(\tan θ + \sec θ - 1) (\tan θ + \sec θ + 1) = \frac{2 \sin θ}{(1 - \sin θ)}

sin θ (1 + tan θ) + c o s θ (1 + cot θ) = csc θ + sec θ

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