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Answer 1

$\huge \rm༆ Answer ༄$

The equivalent expression is ~

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Answer 2

For x such that 0 < x < $\frac{\pi}{2}$, the mathematical expression is:

$\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = 1+1=2$

Given the following data:

• $\frac{\sqrt{1 \;-cos^2x} }{sinx}$

• $\frac{\sqrt{1 \;-sin^2x} }{cosx}$

In Trigonometry, you should take note of the following mathematical expression:

$sin^2x + cos^2x = 1$

Therefore, we can obtain the following:

$sin^2x = 1 - cos^2x$   ...equation 1.

$cos^2x = 1 - sin^2x$    ...equation 2.

Substituting the equations respectively, we have:

$\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = \frac{\sqrt{sin^2x} }{sinx} + \frac{\sqrt{cos^2x} }{cosx}$

Taking the square roots, we have:

$\frac{sinx}{sinx} + \frac{cosx}{cosx} = 1 +1\\\\1+1=2$

Therefore, for x such that 0 < x < $\frac{\pi}{2}$, the expression is:

$\frac{\sqrt{1 \;-cos^2x} }{sinx} + \frac{\sqrt{1 \;-sin^2x} }{cosx} = 1+1=2$

Read more on trigonometry here: brainly.com/question/4515552