If \alpha = \arccos\left(-\frac{5}{13}\right), what is the exact value of \tan\left(\alpha… — ACT

If $\alpha = \arccos\left(-\frac{5}{13}\right)$ , what is the exact value of $\tan\left(\alpha - \frac{\pi}{4}\right)$ ?

The exact value of

\tan\left(\alpha - \frac{\pi}{4}\right)

\frac{17}{7}

Evaluating

\alpha = \arccos\left(-\frac{5}{13}\right)

places

\alpha

in Quadrant II (

\frac{\pi}{2} < \alpha < \pi

). In this quadrant, cosine is negative (

-\frac{5}{13}

) and sine is positive (

\frac{12}{13}

), giving

\tan\alpha = -\frac{12}{5}

. Substituting

\tan\alpha = -\frac{12}{5}

and

\tan\left(\frac{\pi}{4}\right) = 1

into the identity

\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

yields

\frac{-\frac{12}{5} - 1}{1 - \frac{12}{5}} = \frac{-\frac{17}{5}}{-\frac{7}{5}} = \frac{17}{7}

Determine the quadrant and reference values for

\alpha = \arccos\left(-\frac{5}{13}\right)

Since the principal range of

\arccos(x)

[0, \pi]

and

-\frac{5}{13} < 0

\alpha

lies in Quadrant II where

\cos\alpha = -\frac{5}{13}

and

\sin\alpha > 0

Inverse cosine returns angles in Quadrant II for negative arguments.

Calculate

\sin\alpha

and

\tan\alpha

\sin\alpha = \sqrt{1 - \left(-\frac{5}{13}\right)^2} = \sqrt{\frac{144}{169}} = \frac{12}{13}

, so

\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{12/13}{-5/13} = -\frac{12}{5}

Using the Pythagorean identity and definition of tangent.

Apply the tangent angle subtraction identity

\tan\left(\alpha - \frac{\pi}{4}\right) = \frac{\tan\alpha - \tan\left(\frac{\pi}{4}\right)}{1 + \tan\alpha \tan\left(\frac{\pi}{4}\right)} = \frac{-\frac{12}{5} - 1}{1 + \left(-\frac{12}{5}\right)(1)} = \frac{-\frac{17}{5}}{-\frac{7}{5}} = \frac{17}{7}

Evaluating

\tan\left(\frac{\pi}{4}\right) = 1

and simplifying the complex fraction.

Inverse Trigonometric Functions and Compound Angle Identities

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