Welke gelijkheid
is juist ?

(x geen veelvoud van 90°) 		A.  \(\boldsymbol{\tan x = \frac {1} {\cos x}-1 }\)
B.  \(\boldsymbol{\tan x = \frac {1} {\cos^2 x} }\)
C.  \(\boldsymbol{\tan x = 1 - \frac {1} {\cos^2 x} }\)
D.  \(\boldsymbol{\tan^2 x = 1 - \frac {1} {\cos^2 x} }\)
E.  \(\boldsymbol{\tan^2 x = \frac {1} {\cos^2 x} - 1 }\)
A    B    C    D    E 

[ 4-3856 - op net sinds 6.7.07-(E)-2.11.2023 ]

Translation in   E N G L I S H

Which equality
is correct ?

(x not a multiple of 90°) 		A.   \(\boldsymbol{\tan x = \frac {1} {\cos x}-1 }\)
B.   \(\boldsymbol{\tan x = \frac {1} {\cos^2 x} }\)
C.   \(\boldsymbol{\tan x = 1 - \frac {1} {\cos^2 x} }\)
D.   \(\boldsymbol{\tan^2 x = 1 - \frac {1} {\cos^2 x} }\)
E.   \(\boldsymbol{\tan^2 x = \frac {1} {\cos^2 x} - 1 }\)

Oplossing - Solution

Je kan de formule afleiden uit de formule die je vlak na de grondformule geleerd heb, namelijk \(1 + \tan² \alpha =\frac {1} {\cos^2 \alpha} \)
gricha