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Why ?
(xi) Cosine and secant functions are even functions and all other trigonometric functions are odd functions.

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Solution

$A f u n c t i o n f (x) i s s a i d t o b e e v e n i f f (- x) = f (x) A f u n c t i o n f (x) i s s a i d t o b e o d d i f f (- x) = - f (x) \cos (- x) = \cos (π - x - π) = \cos (π - (π + x)) = - \cos (π + x) = - (- \cos x) = \cos x \cos (- x) = \cos x T h e r e f o r e \cos x i s a n e v e n f u n c t i o n s e c (- x) = \frac{1}{\cos (- x)} = \frac{1}{\cos x} = s e c x s e c (- x) = s e c x T h e r e f o r e s e c x i s a n e v e n f u n c t i o n \sin (- x) = \sin (π - x - π) = \sin (π - (π + x)) = \sin (π + x) = - \sin x \sin (- x) = - \sin x T h e r e f o r e \sin x i s a n o d d f u n c t i o n \tan (- x) = \tan (π - x - π) = \tan (π - (π + x)) = \tan (π + x) = - \tan x t a n (- x) = - t a n x T h e r e f o r e \tan x i s a n o d d f u n c t i o n c o t (- x) = \frac{1}{\tan (- x)} = \frac{1}{- \tan x} = - \frac{1}{\tan x} = - c o t x c o t (- x) = - c o t x T h e r e f o r e c o t x i s a n o d d f u n c t i o n$

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