Question

Look at the image. (calculus)

Answer 1

Answer: Choice H)  2

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Explanation:

Recall that the pythagorean trig identity is $\sin^2 x + \cos^2x = 1$

If we were to isolate sine, then,

$\sin^2 x + \cos^2x = 1\\\\\sin^2 x = 1-\cos^2x\\\\\sin x = \sqrt{1-\cos^2x}$

We don't have to worry about the plus minus because sine is positive when 0 < x < pi/2.

Through similar calculations, $\cos x = \sqrt{1-\sin^2x}$

Cosine is also positive in this quadrant.

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So,

$\frac{\sqrt{1-\cos^2x}}{\sin x}+\frac{\sqrt{1-\sin^2x}}{\cos x}\\\\\frac{\sin x}{\sin x}+\frac{\cos x}{\cos x}\\\\1+1\\\\2$

Therefore,

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

is an identity as long as 0 < x < pi/2

Answer 2

Answer:

2

Step-by-step explanation:

$=\frac{\sqrt{1-cos^{2}x }}{sinx} +\frac{\sqrt{1-sin^{2}x } }{cosx} \\\\When \:0 < x < \frac{\pi }{2} \\\\=\frac{\sqrt{sin^{2}x } }{sinx} +\frac{\sqrt{cos^{2}x } }{cosx} \\\\=\frac{sinx}{sinx} +\frac{cosx}{cosx} \\\\=1+1\\\\=2$

Here we use the formula sin²x=1-cos²x and cos²x=1-sin²x.