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Trigonometric Ratios of Compound Angles

Prove that ...

Question

Prove that $tan (\frac{π}{4} + θ) - tan (\frac{π}{4} - θ) = 2 tan 2 θ$

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Solution

$t a n (\frac{π}{4} + θ) - t a n (\frac{π}{4} - θ)$
$= (\frac{1 + t a n θ}{1 - t a n θ}) - (\frac{1 - t a n θ}{1 + t a n θ})$
$= \frac{(1 + t a n θ)^{2} - (1 - t a n θ)^{2}}{(1 - t a n θ) (1 + t a n θ)}$
$= \frac{(1 + t a n^{2} θ + 2 t a n θ) - (1 + t a n^{2} θ - 2 t a n θ)}{1 - t a n^{2} θ}$
$= \frac{4 t a n θ}{1 - t a n^{2} θ}$
$= \frac{2.2 t a n θ}{1 - t a n^{2} θ}$
$= 2 tan 2 θ$

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tan (\frac{π}{4} + θ) - tan (\frac{π}{4} - θ) = 2 tan 2 θ

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tan (\frac{π}{4} + θ) + tan (\frac{π}{4} - θ) = 2 sec 2 θ

tan (π / 4 + θ) \times tan (π / 4 - θ) = 1

tan (\frac{π}{4} + \frac{θ}{2}) = \frac{1}{sec θ - t a n θ}

\frac{tan (\frac{π}{4} + θ) + tan (\frac{π}{4} - θ)}{tan (\frac{π}{4} + θ) - tan (\frac{π}{4} - θ)} =

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