Trigonometric Ratios of (270° - θ)

$ \displaystyle \ \ \ \text{In}\ \vartriangle PON,$

$ \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။


စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!
Previous Post Next Post