Establish each identity below.

(1) 9 sec^2 (x) - 5 tan^2 (x) = 5 + 4 sec^2 (x)

(2) cos(x)/[1 + sin(x)] + [1 + sin(x)]/cos(x) = 2 sec (x)

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- Aug 23rd 2008, 01:47 PMmagentaritaTrigonometric Identities
Establish each identity below.

(1) 9 sec^2 (x) - 5 tan^2 (x) = 5 + 4 sec^2 (x)

(2) cos(x)/[1 + sin(x)] + [1 + sin(x)]/cos(x) = 2 sec (x) - Aug 23rd 2008, 01:53 PMo_O
#1:

Note that $\displaystyle \tan^{2} x = \sec^{2} x - 1$. So substitute it in and simplify.

#2:

Combine into a single fraction:

$\displaystyle \frac{\cos x}{1+\sin x} {\color{blue}\cdot \frac{\cos x}{\cos x} } \: + \: \frac{1+ \sin x}{\cos x} {\color{blue} \cdot \frac{1+ \sin x}{1 + \sin x} }$

$\displaystyle = \frac{\cos^2 x + (1+ \sin x)^2}{\cos x(1+ \sin x)}$

Expand the numerator and simplify.

Post back if you have any troubles! - Aug 23rd 2008, 03:26 PMmagentaritaWhere???
- Aug 23rd 2008, 04:08 PMskeeterQuote:

What in the question indicated that you needed to make cosx/cosx

on the left side and 1 + sinx/1 + sinx on the right side?

- Aug 23rd 2008, 05:51 PMmagentaritaI see..
- Aug 23rd 2008, 06:52 PMo_O
An analogy:

$\displaystyle \frac{1}{2} + \frac{2}{3}$

$\displaystyle = \frac{1}{2} {\color{blue}\cdot \frac{3}{3} } + \frac{2}{3} {\color{blue}\cdot \frac{2}{2}}$

$\displaystyle = \frac{3}{6} + \frac{4}{6}$

$\displaystyle =\frac{7}{6}$ - Aug 24th 2008, 02:48 PMmagentaritaGreat...
Great analogy!

Thanks