Trig identity

**TacticalPro** · March 7th 2011, 05:29 PM

hi i really need help with this homework trig identity it's really angering me. please tell me how you would start it off to get me started thanks.

(secx + tanx) / (secx - tanx) = (1 + 2sinx + sin(x)^2) / cos(x)^2

thanks

**Ackbeet** · March 7th 2011, 05:31 PM

I would definitely start off by writing everything in terms of sines and cosines. Where does that take you?

**Soroban** · March 7th 2011, 08:06 PM

Hello, TacticalPro!

$\displaystyle \frac{\sec x + \tan x}{\sec x - \tan x} \;=\;\frac{1 + 2\sin x + \sin^2\!x}{\cos^2\!x}$

Start with the left side; multiply by $\dfrac{\sec x + \tan x}{\sec x + \tan x}$

. . $\displaystyle \frac{\sec x + \tan x}{\sec x - \tan x}\cdot\frac{\sec x + \tan x}{\sec x + \tan x}$

. . $\displaystyle =\;\frac{(\sec x + \tan x)^2}{\underbrace{\sec^2\!x - \tan^2\!x}_{\text{This is 1}}}$

. . $=\;\sec^2\!x + 2\sec x\tan x + \tan^2\!x$

. . $\displaystyle =\;\frac{1}{\cos^2\!x} + 2\!\cdot\!\frac{1}{\cos x}\!\cdot\!\frac{\sin x}{\cos x} + \frac{\sin^2\!x}{\cos^2\!x}$

. . $\displaystyle =\;\frac{1 + 2\sin x + \sin^2\!x}{\cos^2\!x}$

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