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Sin2A and Cos2A in Terms of tanA

Prove that:ta...

Question

Prove that:

$t a n (\frac{π}{4} + θ) + t a n (\frac{π}{4} - θ) = 2 s e c 2 θ$

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Solution

LHS $= t a n (\frac{π}{4} + θ) + t a n (\frac{π}{4} - θ)$

$= \frac{t a n \frac{π}{4} + t a n θ}{1 - t a n \frac{π}{4} t a n θ} + \frac{t a n \frac{π}{4} - t a n θ}{1 + t a n \frac{π}{4} t a n θ} = \frac{1 + t a n θ}{1 - t a n θ} + \frac{1 - t a n θ}{1 + t a n θ} [∵ t a n \frac{π}{4} = 1] = \frac{(1 + t a n^{2} θ + 2 t a n θ) + (1 + t a n^{2} θ - 2 t a n θ)}{(1 - t a n θ) (1 + t a n θ)} = \frac{2 (1 + t a n^{2} θ)}{1 - t a n^{2} θ}$

$= \frac{2 s e c^{2} θ}{1 - \frac{s i n^{2} θ}{c o s^{2} θ}} [∵ s e c^{2} θ = 1 + t a n^{2} θ] = \frac{2 s e c^{2} θ . c o s^{2} θ}{c o s^{2} θ - s i n^{2} θ} [∵ s e c = \frac{1}{c o s θ}] = \frac{2}{c o s 2 θ} = 2 s e c 2 θ = R H S$

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

csc (π / 4 + θ) csc (π / 4 - θ) = sec (π / 4 + θ) sec (π / 4 - θ) = 2 sec 2 θ

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

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

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

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