I have no idea how to solve these problems, any help wold be aprreciated
a) tanθ - 1 = (sin^2θ - cos^2θ) / (sinθcosθ + cos^2θ)
b) sinx + sinxcot^2x = cscx
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I have no idea how to solve these problems, any help wold be aprreciated
a) tanθ - 1 = (sin^2θ - cos^2θ) / (sinθcosθ + cos^2θ)
b) sinx + sinxcot^2x = cscx
Hello, AdamEdward!
Quote:
$\displaystyle (a)\;\;\tan\theta - 1 \:=\: \frac{\sin^2\!\theta - \cos^2\!\theta}{\sin\theta\cos\theta + \cos^2\!\theta}$
Factor the right side: .$\displaystyle \frac{(\sin\theta - \cos\theta)(\sin\theta + \cos\theta)}{\cos\theta(\sin\theta + \cos\theta)} $
Reduce: .$\displaystyle \frac{\sin\theta - \cos\theta}{\cos\theta} \;\;=\;\;\frac{\sin\theta}{\cos\theta} - \frac{\cos\theta}{\cos\theta} \;\;=\;\; \tan\theta - 1$
Quote:
$\displaystyle (b)\;\;\sin x + \sin x\cot^2\!x \:=\: \csc x$
Factor the left side: .$\displaystyle \sin x\cdot\underbrace{(1 + \cot^2\!x)}_{\text{This is }\csc^2\!x} \;\;=\;\;\sin x\cdot\csc^2\!x \;\;=\;\;\frac{1}{\csc x}\!\cdot \csc^2\!x \;\;=\;\;\csc x$
