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For
x such that
0 < x < (pi)/(2), the expression
(sqrt(1-cos^(2)x))/(sin x)+(sqrt(1-sin^(2)x))/(cos x) is equivalent to:
F. 0
G. 1
H. 2

-tan x

sin 2x

For $x$ such that $0<x<\frac{\pi}{2}$ , the expression $\frac{\sqrt{1-\cos ^{2} x}}{\sin x}+\frac{\sqrt{1-\sin ^{2} x}}{\cos x}$ is equivalent to: $\newline$ F. $0$ $\newline$ G. $1$ $\newline$ H. $2$ $\newline$ $-\tan x$ $\newline$ $\sin 2 x$

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Simplify rational expressions

Full solution

Q. For

x

such that

0<x<\frac{\pi}{2}

, the expression

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

is equivalent to:

\newline

F.

0

\newline

G.

1

\newline

H.

2

\newline

-\tan x

\newline

\sin 2 x

Use Pythagorean Identity: We are given the expression $(\sqrt{1-\cos^{2}x})/(\sin x)+(\sqrt{1-\sin^{2}x})/(\cos x)$ and we need to simplify it. We know that $\sin^2(x) + \cos^2(x) = 1$ , which is the Pythagorean identity. We can use this identity to simplify the square roots in the expression.
Simplify First Term: First, let's simplify the first term $(\sqrt{1-\cos^{2}x})/(\sin x)$ . Using the Pythagorean identity, we can replace $1 - \cos^2(x)$ with $\sin^2(x)$ . This gives us $(\sqrt{\sin^2(x)})/(\sin x)$ .
Simplify Second Term: Since the square root of $\sin^2(x)$ is $|\sin(x)|$ and we know that $0 < x < \frac{\pi}{2}$ , $\sin(x)$ is positive in this interval. Therefore, $\frac{\sqrt{\sin^2(x)}}{\sin x}$ simplifies to $\frac{\sin(x)}{\sin(x)}$ , which equals $1$ .
Combine Simplified Terms: Now, let's simplify the second term $(\sqrt{1-\sin^{2}(x)})/(\cos x)$ . Again, using the Pythagorean identity, we can replace $1 - \sin^2(x)$ with $\cos^2(x)$ . This gives us $(\sqrt{\cos^2(x)})/(\cos x)$ .
Final Result: Since the square root of $\cos^2(x)$ is $|\cos(x)|$ and we know that $0 < x < \frac{\pi}{2}$ , $\cos(x)$ is positive in this interval. Therefore, $\frac{\sqrt{\cos^2(x)}}{\cos x}$ simplifies to $\frac{\cos(x)}{\cos(x)}$ , which equals $1$ .
Final Result: Since the square root of $\cos^2(x)$ is $|\cos(x)|$ and we know that $0 < x < \frac{\pi}{2}$ , $\cos(x)$ is positive in this interval. Therefore, $\frac{\sqrt{\cos^2(x)}}{\cos x}$ simplifies to $\frac{\cos(x)}{\cos(x)}$ , which equals $1$ . Adding the simplified terms from the previous steps, we get $1 + 1$ , which equals $2$ . Therefore, the expression $\frac{\sqrt{1-\cos^{2}x}}{\sin x}+\frac{\sqrt{1-\sin^{2}x}}{\cos x}$ is equivalent to $2$ for $0 < x < \frac{\pi}{2}$ .

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