# Help Simplifying Trig Identities

• Mar 15th 2009, 07:33 AM
Help Simplifying Trig Identities
I need some help simplifying these identities...

#1 5(sec^2x - sin^2xsec^2x)

#2 sin(π/2 - x)(- secx)

and verifying this one...

#1 cscθ/cotθ + tanθ = cosθ
• Mar 15th 2009, 08:38 AM
Soroban

Quote:

I need some help simplifying these expressions.

$\displaystyle (1)\;\;5\left(\sec^2\!x - \sin^2\!x\sec^2\!x\right)$

We have: .$\displaystyle 5\left(\sec^2\!x - \sin^2\!x\underbrace{\sec^2\!x}_{\frac{1}{\cos^2\! x}} \right) \;=\;5\left(\sec^2\!x - \underbrace{\frac{\sin^2\!x}{\cos^2\!x}}_{\tan^2\! x}\right)$

. . . . . . . $\displaystyle = \;5\underbrace{\left(\sec^2\!x - \tan^2\!x\right)}_{\text{This is 1}} \;=\;5\cdot1 \;=\;5$

Quote:

$\displaystyle (2)\;\;\sin\left(\tfrac{\pi}{2}- x\right)(-\sec x)$

We have: .$\displaystyle \underbrace{\sin\left(\tfrac{\pi}{2}-x\right)}_{\text{This is }\cos x}\underbrace{(-\sec x)}_{-\frac{1}{\cos x}} \;=\;\cos x\left(-\frac{1}{\cos x}\right) \;=\;-1$

Quote:

Verify this identity: .$\displaystyle (1)\;\;\frac{\csc\theta}{\cot\theta + \tan\theta} \:= \:\cos\theta$

We have: . $\displaystyle \frac{\dfrac{1}{\sin\theta}} {\dfrac{\cos\theta}{\sin\theta} + \dfrac{\sin\theta}{\cos\theta}}$

Multiply by $\displaystyle \frac{\sin\theta\cos\theta}{\sin\theta\cos\theta}$

. . $\displaystyle \frac{\sin\theta\cos\theta\left(\dfrac{1}{\sin\the ta}\right)} {\sin\theta\cos\theta\left(\dfrac{\cos\theta}{\sin \theta} + \dfrac{\sin\theta}{\cos\theta}\right)} \;=\;\frac{\cos\theta}{\underbrace{\cos^2\!\theta + \sin^2\!\theta}_{\text{This is 1}}} \;=\;\frac{\cos\theta}{1} \;=\;\cos\theta$