$ \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 second quadrant}\text{.}$
$ \displaystyle \therefore {y}'=x\ \text{and }{x}'=-y.$
$ \displaystyle \ \ \ \sin (90{}^\circ +\theta )={y}'=x=\cos \theta $
$ \displaystyle \ \ \ \cos (90{}^\circ +\theta )={x}'=-y=-\sin \theta $
$ \displaystyle \ \ \ \tan (90{}^\circ +\theta )=\frac{{{y}'}}{{{x}'}}=-\frac{x}{y}=\cot \theta $
$ \displaystyle \ \ \ \cot (90{}^\circ +\theta )=\frac{{{x}'}}{{{y}'}}=-\frac{y}{x}=\tan \theta $
$ \displaystyle \ \ \ \sec (90{}^\circ +\theta )=\frac{1}{{{x}'}}=-\frac{1}{y}=-\operatorname{cosec}\theta $
$ \displaystyle \ \ \ \operatorname{cosec}(90{}^\circ +\theta )=\frac{1}{{{y}'}}=\frac{1}{x}=sec\theta $
$ \displaystyle \theta$ တန္ဖိုး႐ိုက္ထည့္ၾကည့္ပါ။
စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!