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Circular Measurement of Angle

Prove that t...

Question

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

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Solution

LHS $2. \frac{1 + {tan}^{2} θ}{1 - {tan}^{2} θ} = 2. \frac{1}{cos 2 θ} = 2. sec 2 θ$

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Similar questions

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

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

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

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

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

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