# Trigonometric Ratios of (180° – θ)

$\displaystyle \begin{array}{l}\ \ \ \text{In}\ \vartriangle PON,\\\\\ \ \ \sin \theta =y\\\\\ \ \ \cos \theta =x\end{array}$

$\displaystyle \ \ \ \tan \theta =\frac{y}{x}$

$\displaystyle \ \ \ \cot \theta =\frac{x}{y}$

$\displaystyle \ \ \ \sec \theta =\frac{1}{x}$

$\displaystyle \ \ \ \operatorname{cosec}\theta =\frac{1}{y}$

$\displaystyle \begin{array}{l}\ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle PON,\\\\\ \ \ {y}'=y\ \text{and }{x}'=x\ \text{numerically}\text{.}\\\\\ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the second quadrant}\text{.}\\\\\therefore {y}'=y\ \text{and }{x}'=-x.\\\\\therefore \sin (180{}^\circ -\theta )={y}'=y=\sin \theta \\\\\ \ \ \cos (180{}^\circ -\theta )={x}'=-x=-\cos \theta \end{array}$

$\displaystyle \ \ \ \tan (180{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{y}{x}=-\tan \theta$

$\displaystyle \ \ \ \cot (180{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{x}{y}=-\cot \theta$

$\displaystyle \ \ \ \sec (180{}^\circ -\theta )=\frac{1}{{{x}'}}=-\frac{1}{x}=-\sec \theta$

$\displaystyle \ \ \ \operatorname{cosec}(180{}^\circ -\theta )=\frac{1}{{{y}'}}=\frac{1}{y}=\operatorname{cosec}\theta$

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