Question

I really need help with these questions:

Answer 1

If we take the Pythagorean identity identity sin^2 x + cos^2 x = 1 then
                          (cos^2 x + sin^2 x) / (cot^2 x - csc^2 x)
The numerator becomes 1 since addition order matters not.
                                                 1 / (cot^2 x - csc^2 x)
If we factor the denominator out a negative
                                                1 / -(csc^2 x - cot^2 x)
Consider sin^2 x + cos^2 x = 1. Divide both sides by sin^2 x to get
                                         1 + cot^2 x = csc^2 x
Subtract both sides by cot^2 x to get 1 = csc^2 x - cot^2 x.
Replace the denominator
                                                 1 / -(1) = -1
For cos^2 θ / sin^2 θ + csc θ sin θ, we use cscθ = 1/sinθ and cosθ/sinθ = cotθ so
                                = cos^2 θ / sin^2 θ + 1
                                = cot^2 θ + 1
We use 1 + cot^2 θ = csc^2 θ to simplify this to
                              = csc^2 θ

Answers:         -1
                          csc^2 θ

Answer 2

1)

$\bf sin^2(\theta)+cos^2(\theta)=1 \qquad \qquad cot(\theta)=\cfrac{cos(\theta )}{sin(\theta)} \qquad csc(\theta)=\cfrac{1}{sin(\theta)}\\\\ -------------------------------\\\\ \cfrac{cos^2(x)+sin^2(x)}{cot^2(x)-csc^2(x)}\implies \cfrac{1}{\frac{cos^2(x)}{sin^2(x)}-\frac{1}{sin^2(x)}}\implies \cfrac{1}{\frac{cos^2(x)-1}{sin^2(x)}} \\\\\\ \cfrac{1}{\frac{-[1-cos^2(x)]}{sin^2(x)}}\implies \cfrac{1}{\frac{-[\underline{sin^2(x)}]}{\underline{sin^2(x)}}}\implies \cfrac{1}{-1}\implies -1$



2)

$\bf \cfrac{cos^2(\theta )}{sin^2(\theta )}+csc(\theta )sin(\theta )\implies \cfrac{cos^2(\theta )}{sin^2(\theta )}+\cfrac{1}{sin(\theta )}\cdot sin(\theta ) \\\\\\ \cfrac{cos^2(\theta )}{sin^2(\theta )}+1\implies \cfrac{cos^2(\theta )+sin^2(\theta )}{sin^2(\theta )}\implies \cfrac{1}{sin^2(\theta )}\implies csc^2(\theta )$