Prove -tan-sin- -tan-sin-1 -1

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Range of Trigonometric Ratios from 0 to 90 Degrees

Prove :tanθ+s...

Question

Prove :

(tanθ + sinθ) / (tanθ - sinθ) = (secθ + 1) / (secθ - 1)

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\frac{1 + s e c θ - t a n θ}{1 + s e c θ + t a n θ} = \frac{1 - s i n θ}{c o s θ}

Prove the following

1. $(1 - {sin}^{2} A) s e c^{2} A = 1$

2. ${sec}^{4} θ - {sec}^{2} θ = {tan}^{4} θ + {tan}^{2}$

3. $(sec θ - tan θ)^{2} = \frac{1 - sin θ}{1 + sin θ}$

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

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