Help Simplifying Trig Identities

Hello, adam456006!

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I need some help simplifying these expressions.

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

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

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

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

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

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

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

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

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