Tangent of Supplementary Angle
Theorem
- $\map \tan {\pi - \theta} = -\tan \theta$
where $\tan$ denotes tangent.
That is, the tangent of an angle is the negative of its supplement.
Proof
| \(\ds \map \tan {\pi - \theta}\) | \(=\) | \(\ds \frac {\map \sin {\pi - \theta} } {\map \cos {\pi - \theta} }\) | Tangent is Sine divided by Cosine | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\sin \theta} {-\cos \theta}\) | Sine of Supplementary Angle and Cosine of Supplementary Angle | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\tan \theta\) |
$\blacksquare$
Also see
- Sine of Supplementary Angle
- Cosine of Supplementary Angle
- Cotangent of Supplementary Angle
- Secant of Supplementary Angle
- Cosecant of Supplementary Angle
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 5$: Trigonometric Functions: Functions of Angles in All Quadrants in terms of those in Quadrant I
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Appendix $12$: Trigonometric formulae: Symmetry
- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $14$: Trigonometric formulae: Symmetry