sec60°−sec0°=

$s e c 60 ° - s e c 0 ° =$

A
$\frac{2}{\sqrt{3}}$
B
$1$
C
not defined
D
$2$

Solution:

\sec 60 ° = 2

(∴ \cos 60 ° = \frac{1}{2})

$\sec 0 ° = 1$ $(∴ \cos 0 ° = 1)$

$∴ \sec 60 ° - \sec 0 ° = 2 - 1 = 1$

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