Question

Struggling in Trig any help is much appreciated, thank you!

Simplify the expression.
csc^2 x-1/1+sinx

Answer 1

`(\csc^2x-1)/(1+\sin x)=(\cot^2x)/(1+\sin x)`

which comes from the fact that
`\sin^2x+\cos^2x=1\implies1+\cot^2x=\csc^2x`.


`(\cot^2x)/(1+\sin x)*(1-\sin x)/(1-\sin x)=(\cot^2x(1-\sin x))/(1-\sin^2x)=(\cot^2x(1-\sin x))/(\cos^2x)`

Now,
`\cot x=(\cos x)/(\sin x)`, which means


`(\cot^2x(1-\sin x))/(\cos^2x)=((\cos^2x)/(\sin^2x)(1-\sin x))/(\cos^2x)=\frac{\frac1{\sin^2x}(1-\sin x)}1=(1-\sin x)/(\sin^2x)`

Recalling that
`\csc x=\frac1{\sin x}`, you can write this as


`\frac1{\sin^2x}-(\sin x)/(\sin^2x)=\csc^2x-\frac1{\sin x}=\csc^2x-\csc x`

Answer 2

Answer: trigonometric identities

1 C sin^2

2 B csc x(csc x-1)

3 D 1-sin theta/ csc theta

4 D tan x sin x

5 A cos x cot2 x