Question

Find an acute angle \(\theta\) that satisfies the equation $$\tan \theta=\cot \left(\theta+45^{\circ}\right)$$

Short answer

\( \theta = \tan^{-1}(-1 + \sqrt{2}) \)

Step-by-step solution

- Recall the Cotangent definition

Recall that the cotangent function is the reciprocal of the tangent function. Therefore, \ \ \ \ \ \ \ \ \ \( \cot(x) = \frac{1}{\tan(x)} \)

- Substitute Cotangent

Substitute the definition of the cotangent function into the equation: \ \( \tan(\theta) = \frac{1}{\tan(\theta + 45^{\circ})}\)

- Use Tangent Addition Formula

Recall the tangent addition formula: \ \( \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \) \ \ Substitute \( a = \theta \) and \( b = 45^{\circ} \): \ \( \tan(\theta + 45^{\circ}) = \frac{\tan(\theta) + 1}{1 - \tan(\theta)} \)

- Substitute Tangent Addition Result

Substitute the result from the tangent addition formula into our equation: \ \( \tan(\theta) = \frac{1}{\frac{\tan(\theta) + 1}{1 - \tan(\theta)}} \) \ Simplify the right side: \ \( \tan(\theta) = \frac{1 - \tan(\theta)}{\tan(\theta) + 1} \)

- Cross Multiply to Simplify

Cross-multiply to eliminate the fraction: \ \( \tan(\theta)(\tan(\theta) + 1) = 1 - \tan(\theta) \) \ \ Expand and simplify: \ \( \tan^2(\theta) + \tan(\theta) = 1 - \tan(\theta) \) \ Moving everything to one side, we get: \ \( \tan^2(\theta) + 2\tan(\theta) - 1 = 0 \)

- Solve the Quadratic Equation

Solve the quadratic equation \( \tan^2(\theta) + 2\tan(\theta) - 1 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = -1 \): \ \ \( x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \) \ \( x = \frac{-2 \pm \sqrt{4 + 4}}{2} \) \ \( x = \frac{-2 \pm \sqrt{8}}{2} \) \ \( x = \frac{-2 \pm 2\sqrt{2}}{2} \) \ \( x = -1 \pm \sqrt{2} \) \ Since \( x = \tan(\theta) \), we consider the positive value for acute angles: \( \tan(\theta) = -1 + \sqrt{2} \).

- Find the Angle

The acute angle \( \theta \) satisfying \( \tan(\theta) = -1 + \sqrt{2} \) is obtained using the arctangent function: \ \( \theta = \tan^{-1}(-1 + \sqrt{2}) \)

Key concepts

Tangent

The tangent function is one of the basic trigonometric functions, relating the angle of a right triangle to the ratio of the length of the opposite side to the adjacent side. Mathematically, it is expressed as \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). This function can take any real value and is periodic, repeating every 180 degrees or \( \pi \) radians. Tangent is positive in the first and third quadrants, and negative in the second and fourth quadrants.

Cotangent

Cotangent is another fundamental trigonometric function, which is the reciprocal of the tangent function. It is defined as \( \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{\text{adjacent}}{\text{opposite}} \). Understanding cotangent helps in solving trigonometric equations and transforming between different trigonometric forms. Cotangent is undefined when tangent is zero, meaning it has vertical asymptotes at angles where tangent crosses the x-axis.

Tangent Addition Formula

To solve problems involving the sum of angles, the tangent addition formula is crucial. It states that \( \tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)} \). This formula allows us to find the tangent of a sum of two angles by knowing the tangent values of the individual angles. It's especially useful in simplifying expressions involving multiple angles.

Arc Tangent

The arc tangent or inverse tangent function, denoted as \( \tan^{-1} \theta \) or \( \arctan(\theta) \), is used to find an angle whose tangent value is given. For example, if \( \tan(\theta) = x \), then \( \theta = \arctan(x) \). This function is essential in finding exact angles from ratio values, particularly in trigonometric equations. Arc tangent typically returns angles in the range \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \).

Solving Quadratic Equations

Quadratic equations are polynomial equations of the form \( ax^2 + bx + c = 0 \). To solve these equations, we can use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula is derived from completing the square method and reliably finds the roots of any quadratic equation when \( a, b, \) and \( c \) are coefficients. In trigonometric contexts, when trigonometric functions are equated to such polynomials, solving them using this formula yields possible values for the functions involved.