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How do you find vertical asymptotes for trig functions?
Only the cofunctions have asymptotes. Because csc x = 1/sin x, csc x has vertical asymptotes whenever the denominator is equal to 0, or whenever sin x = 0, which are the multiples of pi (0,1,2,3,4,...). For sec x, it's 1/cos x, thus cos x = 0, x = pi/2 + pi*n, where n is a counting number (0,1,2,etc...). cot x = cos x/sin x, thus its vertical asymptotes are the same as those of csc x. If the function is transformed, look at the number in front of x (for example, csc (2x), that number would be 2)), and divide the fundamental asymptotes (above) by that number. The vertical asymptotes of csc (2x) would be (pi/2, 2pi/2, 3pi/2, etc...).
H ow do you verify that csc theta tan theta sec theta?
csc[]tan[] = sec[]. L: Change csc[] into one over sin[]. Change tan[] into sin[] over cos[]. R: Change sec[] into one over cos[]. 1/sin[] times sin[]/cos[] = 1/cos[]. L: To multiply 2 fractions, multiply the numerators, and multiply the denominators, and put the numerators' product over the denominators' product. R: Nothing more to do. sin[]/sin[]cos[] = 1/cos[]. L: You have a sin[] on both top and bottom. Cross them off to get a one on the top. 1/cos[] = 1/cos[]. Done. [] is theta. L is the left side of the equation. R is the right side.
Solution for tan x plus cot x divided by sec x csc 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.
How do you simplify sec x cot x?
sec(x)*cot(x) = (1/cos(x))*(cos(x)/sin(x)) = (1/sin(x)) = csc(x)
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
