3 years ago

Help me please in a timed test

Mathematics

1 answer:

kkurt [141]3 years ago

8 0

Answer:

The answer is no solution .

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<h3>Answer: Choice H) 2</h3>

=============================================

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,

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

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