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Soroban · #3 October 7th, 2007, 03:38 PM

Hello, mmgolf!

There are some "tricks" you should know for #3 . . .

Quote:

3. Simplify:

. . $(a)\;\;\frac{1+\cos x}{\sin x} + \frac{\sin x}{1+\cos x}$

Multiply the second fraction by: $\frac{1-\cos x}{1-\cos x}$

. . We have: . $\frac{\sin x}{1 + \cos x}\cdot\frac{1-\cos x}{1-\cos x} \;=\;\frac{\sin x(1 - \cos x)}{1 - \cos^2\!x} \;=\;\frac{\sin x(1 - \cos x)}{\sin^2\!x} \;=\;\frac{1-\cos x}{\sin x}$

The problem becomes: . $\frac{1+\cos x}{\sin x} + \frac{1-\cos x}{\sin x}\;=\; \frac{2}{\sin x} \;=\;2\csc x$

Quote:

$(b)\;\;\frac{1+\sec x }{\sin x + \tan x}$

Multiply by: $\frac{\cos x}{\cos x}$

. . $\frac{\cos x}{\cos x}\,\cdot\,\frac{1 + \sec x}{\sin x + \tan x} \;=\;\frac{\cos x + 1}{\sin x\cos x + \sin x} \;=\;\frac{\cos x + 1}{\sin x(\cos x + 1)} \;=\;\frac{1}{\sin x} \;=\;\csc x$

Quote:

$(c)\;\frac{1 + \sin y}{1 + \csc y}$

Multiply by: $\frac{\sin y}{\sin y}$

. . $\frac{\sin y}{\sin y}\cdot\frac{1 + \sin y}{1 + \csc y} \;=\;\frac{\sin y(1 + \sin y)}{\sin y + 1} \;=\;\sin y$