Tan - 4 - 2 - 1-sin- 1-sin- -tan- -

LHS = 1+tan( θ /2 ) #160; #160; 1 tan( θ /2 ) #160; #160; =cos( θ /2 ) #160; +sin( θ /2 ) #160; #160; cos( θ /2 ) #160; sin( θ /2 ) ...

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Trigonometric Ratios of Compound Angles

tan π 4 +θ 2 ...

Question

$tan (\frac{π}{4} + \frac{θ}{2}) = \sqrt{(\frac{1 + sin θ}{1 - sin θ})} = tan θ + sec θ$

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Solution

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tan (\frac{π}{4} + \frac{θ}{2}) = \sqrt{\frac{1 + sin θ}{1 - sin 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. $\frac{tan θ + sec θ - 1}{tan θ - sec θ +} = \frac{1 + sin θ}{cos θ}$

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

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

{sec}^{2} θ = 1 + {tan}^{2} θ

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

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