ANGLES IN QUADRANTS

  ANGLES IN QUADRANTS
  If \(\theta=90^0\)
Let \(op = r = 1{U}\)
 op will coincide with Y-axis     
\(\therefore \,\,op = y = 1 = r\)
x = 0
\(\therefore sin{90^0} = \frac{y}{r} = \frac{1}{1} = 1\,\,\,\,\, \Rightarrow \sin {90^0} = 1\)

\(\cos {90^0} = \frac{x}{r} = \frac{0}{1} = 0\,\,\,\,\, \Rightarrow \cos {90^0} = 0\)
\(\tan {90^0} = \,\,\,\frac{y}{x}\,\,\, = \,\,\,\frac{1}{0}\,\,\, = \infty \,\, \Rightarrow \tan {90^0} = \infty \)
\(\csc {90^0} = \frac{r}{y} = \frac{1}{1} = 1 \Rightarrow \csc {90^0} = 1\) 
\(\sec {90^0} = \frac{r}{x} = \frac{1}{0} = \alpha \Rightarrow \sec {90^0} = \alpha \)   
\(\cot {90^0} = \frac{x}{y} = \frac{0}{1} = 0 \Rightarrow \cot {90^0} = 0\) 
 For \(\theta = {45^0}\) , Trigonometric ratios are defined from Isoscles right triangle .  
  
 Let AB=BC=aU
    \(A{C^2} = A{B^2} + B{C^2} = {a^2} + {a^2}\)
    \(A{C^2} = 2{a^2}\)    
    \(AC = \sqrt 2 aU\)
Let  be Isosceles right angle triangle such that 
\(\therefore sin{45^0} = \frac{{AB}}{{AC}} = \frac{a}{{a\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\,\,\,\,\, \Rightarrow \sin {45^0} = \frac{1}{{\sqrt 2 }}\)
\(\cos {45^0} = \frac{{BC}}{{AC}} = \frac{a}{{a\sqrt 2 }} = \frac{1}{{\sqrt 2 }}\,\,\,\,\, \Rightarrow \cos {45^0} = \frac{1}{{\sqrt 2 }}\)
\(\tan {45^0} = \frac{{AB}}{{AC}} = \frac{a}{a} = 1 \Rightarrow \,\,\,\,\,\tan {45^0} = 1\) 
\(\csc {45^0} = \frac{{AC}}{{AB}} = \frac{{a\sqrt 2 }}{a} = \sqrt 2 \Rightarrow \csc {45^0} = \sqrt 2 \)
\(\sec {45^0} = \frac{{AB}}{{BC}} = \frac{{a\sqrt 2 }}{a} = \sqrt 2 \Rightarrow \sec {45^0} = \sqrt 2 \)
 \(\cot {45^0} = \frac{{BC}}{{AB}} = \frac{a}{a} = 1 \Rightarrow \cot {45^0} = 1\)