Question

Verify the identity.
\[\cot \theta \times \sec \theta = \csc \theta\]

Please explain to me HOW you get the answer that you got, in hopes to understand.

Answer 1

\[cot(\theta)\times sec(\theta)=csc(\theta)\]We know that\[cot(\theta)=\frac{cos(\theta)}{sin(\theta)}\\sec(\theta)=\frac{1}{cos(\theta)}\\csc(\theta)=\frac{1}{sin(\theta)}\]Replacing:\[\frac{cos(\theta)}{sin(\theta)}\times\frac{1}{cos(\theta)}=\frac{1}{sin(\theta)}\]So\[\frac{\not{cos(\theta)}}{sin(\theta)}\times\frac{1}{\not{cos(\theta)}}=\frac{1}{sin(\theta)}\]We've got \[\frac{1}{sin(\theta)}=\frac{1}{sin(\theta)}\Leftrightarrow csc(\theta)=csc(\theta)\]

Answer 2

where did \[⇔ \csc(θ)=\csc(θ)\] come from And I forgot what csc and sec mean

Answer 3

That came from the fact that

\[\large \frac{1}{\sin(x)} = \csc(x)\]

csc = co-secant
sec = secant

Answer 4

@johnweldon1993 thank you very much and thank you too @D3xt3R

Answer 5

You're welcome ;)