For x such that 0 < x < [tex]\frac{\pi}{2}[/tex], the mathematical expression is:
[tex]\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = 1+1=2[/tex]
Given the following data:
- [tex]\frac{\sqrt{1 \;-cos^2x} }{sinx}[/tex]
- [tex]\frac{\sqrt{1 \;-sin^2x} }{cosx}[/tex]
In Trigonometry, you should take note of the following mathematical expression:
[tex]sin^2x + cos^2x = 1[/tex]
Therefore, we can obtain the following:
[tex]sin^2x = 1 - cos^2x[/tex] ...equation 1.
[tex]cos^2x = 1 - sin^2x[/tex] ...equation 2.
Substituting the equations respectively, we have:
[tex]\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = \frac{\sqrt{sin^2x} }{sinx} + \frac{\sqrt{cos^2x} }{cosx}\\\\[/tex]
Taking the square roots, we have:
[tex]\frac{sinx}{sinx} + \frac{cosx}{cosx} = 1 +1\\\\1+1=2[/tex]
Therefore, for x such that 0 < x < [tex]\frac{\pi}{2}[/tex], the expression is:
[tex]\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = 1+1=2[/tex]
Read more on trigonometry here: https://brainly.com/question/4515552
