Trig Identities Help

• May 3rd 2009, 04:54 PM
College Student
Trig Identities Help
1. tan(x+(pi/2)) = -cotx
2. cos^2(x) = csc^2(x)-cot^2(x) / sec^2(x)
3. cosx + cosxtan^2(x) = secx

I'm just not good with this stuff. I tried expanding the left side of the first one using sum and difference formulas and got no where with it, and for the other two I just got everything in terms of sin and cos and got even more lost. Any help is greatly appreciated.
• May 3rd 2009, 08:17 PM
Soroban
Hello, College Student!

Quote:

$1)\;\;\tan\left(x+\tfrac{\pi}{2}\right) \:=\: -\cot x$
$\tan\left(x+\tfrac{\pi}{2}\right) \;=\;\frac{\sin(x + \frac{\pi}{2})}{\cos(x + \frac{\pi}{2})}\;=\;\frac{\sin x\cos\frac{\pi}{2} + \cos x\sin\frac{\pi}{2}} {\cos x\cos\frac{\pi}{2} - \sin x\sin\frac{\pi}{2}}\;=$ . $\frac{\sin x\cdot 0 + \cos x\cdot1}{\cos x\cdot0 - \sin x\cdot1} \;=\;\frac{\cos x}{\text{-}\sin x} \;=\;-\cot x$

Quote:

$2)\;\;\cos^2\!x \:= \:\frac{\csc^2\!x-\cot^2\!x}{\sec^2\!x}$
$\text{On the right: }\;\frac{\overbrace{\csc^2\!x - \cot^2\!x}^{\text{This is 1}}}{\sec^2\!x} \;=\;\frac{1}{\sec^2\!x} \;=\;\cos^2\!x$

Quote:

$3)\;\;\cos x + \cos x\tan^2\!x \:=\: \sec x$
$\text{Factor: }\cos x\overbrace{(1 + \tan^2\!x)}^{\text{This is }\sec^2\!x} \;=\;\cos x\!\cdot\!\sec^2\!x \;=\;\overbrace{(\cos x\!\cdot\!\sec x)}^{\text{This is 1}}\sec x \;=\;\sec x$