In algebra and trigonometry we can have various functions such as sin, cosine , tan and sec and to solve trigonometric equations we should know relation between them .
sec x = 1 / cos x.
tan x = sin x/ cos x.
(1- sinx )/ cos x.
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It generally helps, to manipulate such expressions, if you write everything in terms of sines and cosines.
Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.
Cos x = 1 / Sec x so 1 / Cos x = Sec x Then Tan x = Sin x / Cos x = Sin x * (1 / Cos x) = Sin x * Sec x
Rewrite sec x as 1/cos x. Then, sec x sin x = (1/cos x)(sin x) = sin x/cos x. By definition, this is equal to tan x.
Given y = tan x: dy/dx = sec^2 x(secant of x squared)
f'(x) = 1/tan(x) * sec^2(x) where * means multiply and ^ means to the power of. = cot(x) * sec^2(x) f''(x) = f'(cot(x)*sec^2(x) + cot(x)*f'[sec^2(x)] = -csc^2(x)*sec^2(x) + cot(x)*2tan(x)sec^2(x) = sec^2(x) [cot(x)-csc^2(x)] +2tan(x)cot(x) = sec^2(x) [cot(x)-csc^2(x)] +2