$ \displaystyle \ \ \ \sin \theta =y$
$ \displaystyle \ \ \ \cos \theta =x$
$ \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 \ \ \ \text{Since}\ \vartriangle {P}'O{N}'\cong \vartriangle OPN,$
$ \displaystyle \ \ \ {y}'=x\ \text{and }{x}'=y\ \text{numerically}\text{.}$
$ \displaystyle \ \ \ \text{But }{P}'({x}',{y}')\ \text{lies in the third quadrant}\text{.}$
$ \displaystyle \therefore {y}'=-x\ \text{and }{x}'=-y.$
$ \displaystyle \ \ \ \sin (270{}^\circ -\theta )={y}'=-x=-\cos \theta $
$ \displaystyle \ \ \ \cos (270{}^\circ -\theta )={x}'=-y=-\sin \theta $
$ \displaystyle \ \ \ \tan (270{}^\circ -\theta )=\frac{{{y}'}}{{{x}'}}=\frac{{-x}}{{-y}}=\cot \theta $
$ \displaystyle \ \ \ \cot (270{}^\circ -\theta )=\frac{{{x}'}}{{{y}'}}=\frac{{-y}}{{-x}}=\tan \theta $
$ \displaystyle \ \ \ \sec (270{}^\circ -\theta )=\frac{1}{{{x}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta $
$ \displaystyle \ \ \ \operatorname{cosec}(270{}^\circ -\theta )=\frac{1}{{{y}'}}=-\frac{1}{x}=-\sec\theta $
$ \displaystyle \theta$ တန္ဖိုး ရိုက္ထည့္ပါ။
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