The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
Given y = tan x: dy/dx = sec^2 x(secant of x squared)
tan(sqrtX) + C
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The graph is a circle with a radius of 6, centered at the origin
Yes. Both expressions are the same.
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cot2x-tan2x=(cot x -tan x)(cot x + tan x) =0 so either cot x - tan x = 0 or cot x + tan x =0 1) cot x = tan x => 1 / tan x = tan x => tan2x = 1 => tan x = 1 ou tan x = -1 x = pi/4 or x = -pi /4 2) cot x + tan x =0 => 1 / tan x = -tan x => tan2x = -1 if you know about complex number then infinity is the solution to this equation, if not there's no solution in real numbers.
The graph is a circle with a radius of 6, centered at the origin.
The Answer is 1 coz, 1-Tan squarex = Cot square X. So cot square x divided cot square x is equal to 1
Given y = tan x: dy/dx = sec^2 x(secant of x squared)
To shift a funcion (or its graph) down "a" units, you subtract "a" from the function. For example, x squared gives you a certain graph; "x squared minus a" will give you the same graph, but shifted down "a" units. Similarly, you can shift a graph upwards "a" units, by adding "a" to the function.
A parabola is a type of graph that is not linear, and mostly curved. A parabola has the "x squared" sign in it's equation. A parabola is not only curved, but all the symmetrical. The symmetrical point, the middle of the parabola is called the vertex. You can graph this graph with the vertex, x-intercepts and a y-intercept. A parabola that has a positive x squared would be a smile parabola, and the one with the negative x squared would be a frown parabola. Also, there are the parabolas that are not up or down, but sideways Those parabolas have x=y squared, instead of y = x squared.
the x-axis... obviously! the x-axis... obviously!
tan(sqrtX) + C
Note that for sec²(x) - tan²(x) = 1, we have: -tan²(x) = 1 - sec²(x) tan²(x) = sec²(x) - 1 Rewrite the expression as: ∫ (sec²(x) - 1) dx = ∫ sec²(x) dx - ∫ 1 dx Finally, integrate each expression to get: tan(x) - x + K where K is the arbitrary constant
Let x = theta, since it's easier to type, and is essentially the same variable. Since tan^2(x)=tan(x), you know that tan(x) must either be 1 or zero for this statement to be true. So let tan(x)=0, and solve on your calculator by taking the inverse. Similarly for, tan(x)=1