Arctan Equations Quiz: Solving Equations Using Arctan

tan(arctan(0.5)) = 0.5 because tan and arctan are inverse functions. Then arcsin(0.5) is needed. sin(pi/6) = 0.5 and pi/6 lies within the principal range of arcsin, so arcsin(0.5) = pi/6. The full expression evaluates to pi/6. Option A gives pi/4, which would require arcsin(sqrt(2)/2). Option B gives pi/3, which would require arcsin(sqrt(3)/2). Option D gives pi/2, which would require arcsin(1).

tan(-pi/4) = -1 and -pi/4 lies within the open principal range from -pi/2 to pi/2, so arctan(-1) = -pi/4. Option A gives pi/4, which corresponds to arctan(1), a positive input. Option C gives pi/3, whose tangent is positive sqrt(3). Option D gives -pi/3, whose tangent is -sqrt(3), not -1. This result also follows directly from the odd function property since arctan(1) = pi/4 implies arctan(-1) = -pi/4.

arctan(x) = pi/4 means the angle whose tangent is x equals pi/4. Applying tan to both sides gives x = tan(pi/4) = 1. Option A gives sqrt(3), which satisfies tan(theta) = sqrt(3), meaning arctan(sqrt(3)) = pi/3, not pi/4. Option B gives 1/2, which does not correspond to any standard arctan value at pi/4. Option C gives sqrt(3)/3, which satisfies arctan(sqrt(3)/3) = pi/6, not pi/4.

arctan is defined for every real number with no domain restrictions, confirming A. Since arctan(-x) = -arctan(x) for all x the function is odd, confirming B. Option C is false — the range is the open interval from -pi/2 to pi/2, not from -pi to pi. Option D is false — arctan is not periodic and has no repeating cycle.

tan(-pi/3) = -sqrt(3) and -pi/3 lies within the open principal range from -pi/2 to pi/2, so arctan(-sqrt(3)) = -pi/3. Option A gives -pi/6, whose tangent is -1/sqrt(3), not -sqrt(3). Option B gives -pi/4, whose tangent is -1. Option D gives -pi/2, which is never returned by arctan because tangent is undefined at that angle and the open range excludes that endpoint.

The answer is True. tan(-pi/4) = -1 and -pi/4 lies within the open principal range from -pi/2 to pi/2. Both conditions of the arctan definition are therefore satisfied. This result also follows from the odd function property of arctan: since arctan(1) = pi/4, symmetry requires arctan(-1) = -pi/4. The answer is confirmed both by direct evaluation and by the odd function identity.

The answer is True. arctan(x) returns the unique angle theta in the open interval from -pi/2 to pi/2 satisfying tan(theta) = x. Applying tan to that angle immediately recovers x. Since arctan is defined for every real number and tan undoes arctan exactly on the principal range, this identity holds for every real value of x without any exceptions or domain restrictions.

tan(x) is undefined at x = -pi/2 and x = pi/2 because cos(x) = 0 at those points, causing division by zero in the definition of tangent. Since arctan reverses the tangent function, it can never return an angle where tangent itself does not exist. This is why the principal range of arctan is an open interval, with the boundary values approached as limits but never actually attained.

The range of arctan is the open interval from -pi/2 to pi/2. The endpoints are excluded because y = pi/2 and y = -pi/2 are horizontal asymptotes that the graph approaches but never reaches. Option A is too wide. Option C describes the range of arccos. Option D covers only a quarter of the actual range. arctan approaches but never attains the boundary values, making the interval strictly open.

The answer is True. The tangent function is one-to-one on the open interval from -pi/2 to pi/2, which allows a proper inverse to be defined. The resulting function, arctan, accepts every real number as input because tangent produces every real number as output on that restricted interval. arctan returns the unique angle in the open interval from -pi/2 to pi/2 corresponding to any real input x.

tan(pi/3) = sqrt(3) and pi/3 lies within the open principal range from -pi/2 to pi/2, so arctan(sqrt(3)) = pi/3. Option A gives pi/6, whose tangent is 1/sqrt(3) = sqrt(3)/3, not sqrt(3). Option B gives pi/4, whose tangent is 1. Option C gives pi/2, which is never returned by arctan since tangent is undefined there and the range excludes that endpoint.

tan(pi/4) = 1 and pi/4 is in the principal range, confirming A. Option B is false because tan(-pi/3) = -sqrt(3), not -sqrt(3)/2, so arctan(-sqrt(3)/2) does not equal -pi/3. Option C is false because arctan has no period whatsoever. Since arctan is an odd function every negative input produces a negative output, confirming D.

tan(0) = 0 and 0 lies within the open principal range from -pi/2 to pi/2, so arctan(0) = 0. Option A gives pi/4, which is the solution to tan(theta) = 1. Option B gives -pi/4, the solution to tan(theta) = -1. Option C gives pi/2, which is never returned by arctan because tangent is undefined at that angle and the range is an open interval that excludes pi/2.

arctan returns the unique angle in the open principal range whose tangent equals x, confirming A and B. arctan accepts all real numbers as input with no restrictions, confirming D. Option C is false — the closed interval from 0 to pi describes the range of arccos, not arctan.

tan(pi/4) = 1 and pi/4 lies within the open principal range from -pi/2 to pi/2, so arctan(1) = pi/4. Option A gives -pi/4, which corresponds to tan(-pi/4) = -1, a negative value. Option B gives pi/3, whose tangent is sqrt(3), not 1. Option C gives pi/2, which lies on the boundary of the open principal range and is never returned by arctan since tangent is undefined at that angle.

arctan accepts every real number as input with no domain restrictions, confirming A. The range is the open interval from -pi/2 to pi/2 where boundary values are never reached, confirming B. Option C is false — arctan has no vertical asymptotes because it is defined for all real x. tan(pi/4) = 1 and pi/4 is in the principal range, confirming D.

arctan is the inverse of tangent by definition, confirming A. It returns the unique angle in the open principal range satisfying tan(theta) = x, confirming B. Since arctan is an odd function every negative input produces a negative output, confirming D. Option C is false because arctan(0) = 0 and all negative inputs give negative outputs.

The answer is True. The derivative of arctan(x) is 1 divided by (1 plus x squared), which is always positive for every real x. A consistently positive derivative confirms the function never decreases anywhere on its domain. Unlike sine and cosine, arctan has no intervals where it turns around or oscillates. It increases steadily from negative pi/2 toward positive pi/2 as x increases from negative infinity to positive infinity.

The tangent function outputs every real number across each of its defined intervals, so every real number is a valid input for arctan. Unlike arcsin and arccos, arctan has no restriction on its domain. Option A describes the domain of arcsin. Option B is the range of arctan, not the domain. Option C wrongly excludes all negative inputs. arctan accepts any real value of x with no upper or lower limit.

arctan(-sqrt(3)) asks for the unique angle in the open interval from -pi/2 to pi/2 whose tangent equals -sqrt(3). tan(-pi/3) = -sqrt(3) and -pi/3 lies within the principal range, so theta = -pi/3. Option A gives pi/3, which produces a positive tangent value. Option B gives -pi/6, whose tangent is -1/sqrt(3). Option C gives pi/6, which also produces a positive tangent. Only -pi/3 is correct.

The answer is True. arctan(x) returns the unique angle theta in the open interval from -pi/2 to pi/2 such that tan(theta) = x. Since the domain of arctan is all real numbers, applying tan to the result always recovers x. This identity holds without restriction for every real value of x, confirming that tan and arctan are true inverse functions on their respective domains.