To simplify this sort of things, it helps if, first of all, you convert everything to sines and cosines.
cos x cot x + tan x (original equation)
= cos (cos x / sin x) + (sin x / cos x) (convert to sin and cos)
= cos2x / sin x + sin x / cos x (multiplying in the first term)
= (sin x cos2x + sin x cos x) / sin x cos x (converting common denominator)
= (sin x cos x) (cos x + 1) / (sin x cos x) (factoring the numerator)
= cos x + 1 (cancelling factors in numerator and denominator)
== cot(x)== 1/tan(x) = cos(x)/sin(x) Now substitute cos(x)/sin(x) into the expression, in place of cot(x) So now: sin(x) cot(x) cos(x) = sin(x) cos(x) (cos(x)/sin(x) ) sin(x) cos(x) cos(x)/sin(x) The two sin(x) cancel, leaving you with cos(x) cos(x) Which is the same as cos2(x) So: sin(x) cot(x) cos(x) = cos2(x) ===
y = sec(x)*cot(x)*cos(x)To solve this trigonometric equation, you need to know these identities:sec(x) = 1/(cos(x))cot(x) = 1/(tan(x)) = (cos(x))/(sin(x))Now substitute these identities into the original equation:y = (1/cos(x))*((cos(x))/(sin(x)))*cos(x)Now cancel out the terms that are similar in the numerator and denominator to leave you with:y = (1/(sin(x)))*cos(x)y = (cos(x))/(sin(x))From the aforementioned known identity, the final simplified trigonometric equation becomes:y = cot(x)
(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.
sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)
yes 1 + cot x^2 = csc x^2
cot 70 + 4 cos 70 = cos 70 / sin 70 + 4 cos 70 = cos 70 (1/sin 70 + 4) = cos 70 (csc 70 + 4) Numerical answer varies, depending on whether 70 is in degrees, radians, or grads.
cos*cot + sin = cos*cos/sin + sin = cos2/sin + sin = (cos2 + sin2)/sin = 1/sin = cosec
2 cot(x) + 1 = -1 2 cot(x) = -2 cot(x) = -1 cos(x)/sin(x) = -1 cos(x) = - sin(x) x = 135°, 315°, 495°, ... another one every 180 degrees
Manipulate normally, noting:cot x = cos x / sin xcos² x + sin² x = 1 → sin²x = 1 - cos² xa² - b² = (a + b)(a - b)1 = 1²ab = baa/(bc) = a/b/c(1 + cot x)² - 2 cot x = 1² + 2 cot x + cot² x - 2 cot x= 1 + cot² x= 1 + (cos x / sin x)²= 1 + cos² x / sin² x= 1 + cos² x / (1 - cos² x)= ((1 - cos² x) + cos² x)/(1 - cos² x)= 1/(1² - cos² x)= 1/((1 + cos x)(1 - cos x))= 1/(1 - cos x)/(1 + cos x)QED.
1/ Tan = 1/ (Sin/Cos) = Cos/Sin = Cot (Cotangent)
Without an "equals" sign somewhere, no question has been asked,so there's nothing there that needs an answer.Is it the sum that you're looking for ?csc(x) + cot(x) = 1/sin(x) + cos(x)/sin(x) = [1 + cos(x)] / sin(x)
Suppose csc(x)*sin(x) = cos(x)*cot(x) + y then, ince csc(x) = 1/sin(x), and cot(x) = cos(x)/sin(x), 1 = cos(x)*cos(x)/sin(x) + y so y = 1 - cos2(x)/sin(x) = 1 - [1 - sin2(x)]/sin(x) = [sin2(x) + sin(x) - 1]/sin(x)
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
cot x = (cos x) / (sin x) cos (x - 180) = cos x cos 180 + sin x sin 180 = - cos x sin (x - 180) = sin x cos 180 - cos x sin 180 = - sin x cot (x - 180) = (cos (x - 180)) / (sin (x - 180)) = (- cos x) / (- sin x) = (cos x) / (sin x) = cot x
The TI-83 does not have the cot button, however, if you type 1/tan( then this will work the same as the cot since cot=1/tan. The other way to do this is to type (cos(x))/(sin(x)) where x is the angle you're looking for. This works because cot=cos/sin
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
It just simplifies down to 1=1. You have to use your trig identities... tan=sin/cos cot=cos/sin thus tan x cot= (sin/cos) (cos/sin) since sin is in the numerator for tan, when it is multiplied by cot (which has sin in the denominator) both of the signs cancel and both now have a value of 1. The same happens with cos. so you get 1 x 1=1 so there is your answer. just learn your trig identities and you will understand