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Look at the image. (calculus)

Look at the image. (calculus)-example-1
User TauWich
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2 Answers

7 votes

Answer:

2

Step-by-step explanation:


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

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

User Marko Vranjkovic
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8.8k points
5 votes

Answer: Choice H) 2

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

Step-by-step 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 = √(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 = √(1-\sin^2x)\\\\

Cosine is also positive in this quadrant.

-------------

So,


(√(1-\cos^2x))/(\sin x)+(√(1-\sin^2x))/(\cos x)\\\\(\sin x)/(\sin x)+(\cos x)/(\cos x)\\\\1+1\\\\2

Therefore,


(√(1-\cos^2x))/(\sin x)+(√(1-\sin^2x))/(\cos x)=2

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

User Delinear
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