Solve for -i 1-sin-1-sin-tan-2ii 1-cos-1-cos-2

(i) 1 sinθ #160; #160; 1+sinθ #160; #160; = (1 sinθ #160; ) ^ 2 1 sin ^ 2 θ #160; #160; = 1 2sinθ #160; +sin ^ 2 θ ...

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Circular Measurement of Angle

Solve for θ...

Question

Solve for $θ$ :
i) $\frac{1 - sin θ}{1 + sin θ} = (sec θ - tan θ)^{2}$
ii) $\frac{1 + cos θ}{1 - cos θ} = (cosec θ + cot θ)^{2}$

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\frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} = \frac{2 {sec}^{2} θ}{{tan}^{2} θ - 1}

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Prove the following trigonometric identities:

Prove that:

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

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

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

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

\frac{1 - sinθ}{1 + sinθ} = (secθ - tanθ)^{2}

\frac{1 + cosθ}{1 - cosθ} = (cosecθ + cotθ)^{2}

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