Question

Struggling in Trig any help is much appreciated, thank you!
Simplify the expression.
csc^2 x-1/1+sinx

Answer 1

$\dfrac{\csc^2x-1}{1+\sin x}=\dfrac{\cot^2x}{1+\sin x}$

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

$\dfrac{\cot^2x}{1+\sin x}\times\dfrac{1-\sin x}{1-\sin x}=\dfrac{\cot^2x(1-\sin x)}{1-\sin^2x}=\dfrac{\cot^2x(1-\sin x)}{\cos^2x}$

Now, $\cot x=\dfrac{\cos x}{\sin x}$, which means

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

Recalling that $\csc x=\dfrac1{\sin x}$, you can write this as

$\dfrac1{\sin^2x}-\dfrac{\sin x}{\sin^2x}=\csc^2x-\dfrac1{\sin x}=\csc^2x-\csc x$

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