Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

tan 20°.tan 40°.tan 60°.tan 80° is gelijk aan Gebruik eventueel de formules $\\\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha.\tan\beta}\\\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha.\tan\beta}\\\tan3\alpha=\frac{3\tan\alpha-\tan^3\alpha}{1-3\tan^2\alpha}$	A. tan 50°
	B. tan 100°
	C. 3
	D.
	E.

[ 5-7387 - op net sinds 3.2.15-(E)-20.7.2024 ]

Translation in E N G L I S H

tan 20°.tan 40°.tan 60°.tan 80° =	A. tan 50°
	B. tan 100°
	C.
	D. 3
	E.

Oplossing - Solution

$\\\ tan20^\circ.\tan40^\circ.\tan60^\circ.\tan80^\circ\\ =\tan20^\circ.\tan(60^\circ-20^\circ).\tan60^\circ.\tan(60^\circ+20^\circ)\\ =\sqrt3.\tan20^\circ.\frac{\tan60^\circ-\tan20^\circ}{1+\tan60^\circ.\tan20^\circ}\cdot\frac{\tan60^\circ+\tan20^\circ}{1-\tan60^\circ.\tan20^\circ}\\ =\sqrt3\cdot\tan20^\circ\frac{\tan^260^\circ-\tan^220^\circ}{1-\tan^260^\circ.\tan^220^\circ}\\ =\sqrt3\cdot\frac{3\tan20^\circ-\tan^320^\circ}{1-3.\tan^220^\circ}\\ =\sqrt3\cdot\tan 60^\circ =\sqrt3\cdot\sqrt3=3$