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How do you solve Sin x sec x equals tan x?
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
Sec - cos equals tansin?
Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D
How do you prove that the derivative of sec x is equal to sec x tan x?
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.
How do you identify sec x sin x equals tan 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.
Find the principal solution of sec x equals 2?
sec(x) = 2 so cos(x) = 1/2 and so x = pi/3
How do you find sec x cos x?
sec x = 1/cos x so sec x * cos x = 1
What is sec x cos x?
sec x = 1/cos x sec x cos x = [1/cos x] [cos x] = 1
Is cos 2 x sec x equals 2 cos x - sec x an identity?
Yes, it is. the basic identity is for a double angle relation: cos 2x = 2 cosx cos x -1 since sec x =1/cos x if we multiply both sides by sec x we get cos2xsec x = 2cosxcos x/cos x -1/cos x = 2cos x - sec x
How do you put sec cubed in a calculator?
sec x = 1/cos x → sec³ x = 1/cos³ x or sec³ x = (cos x)^-3 Therefore to enter sec³ x on a calculator: Newer, "natural" calculators: mathio: sec³ x → [x-power] [cos] [<angle>] [)] [navigate →] [(-)] [3] [=] lineio: sec³ x → [(] [cos] [)] [)] [x-power] [(-)] [3] [)] [=] Older, function acts on displayed number calculators: sec³ x → [angle] [cos] [x-power] [3] [±] [=]
How do you solve Sin x sec x equals tan x?
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
Sec - cos equals tansin?
Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D
How do you prove that the derivative of sec x is equal to sec x tan x?
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.
How do you identify sec x sin x equals tan 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.
Find the principal solution of sec x equals 2?
sec(x) = 2 so cos(x) = 1/2 and so x = pi/3
How do you simplify sec x cot x?
sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)
How do you prove tan x plus tan x sec 2x equals tan 2x?
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
What is the reciprocal of a cosine?
1/cos(x)=sec(x). sec is short for secant.
