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Question Number 209781 by alusto22 last updated on 21/Jul/24

\frac{1 + \sin 40 ° - \cos 40 °}{1 + \sin 40 ° + \cos 40 °} = ?

Answered by efronzo1 last updated on 21/Jul/24

\frac{1 + \sin 40 ° - (1 - 2 \sin^{2} 20 °)}{1 + \sin 40 ° + 2 \cos^{2} 20 - 1}

= \frac{2 \sin 20 ° (\cos 20 ° + \sin 20 °)}{2 \cos 20 ° (\cos 20 ° + \sin 20 °)}

= \tan 20 °

Answered by Frix last updated on 22/Jul/24

\frac{1 + \sin 2 x - \cos 2 x}{1 + \sin 2 x + \cos 2 x} = \dots

= \frac{2 \sin x \cos x + {2 \sin}^{2} x}{2 \sin x \cos x + {2 \cos}^{2} x} =

= \frac{\sin x}{\cos x} = \tan x

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