Prove Trigonometric Identities

[nycmath]

See attachment.

[Guest]

Hi, it's Bob...

[tex]\dfrac{\sin B}{1+\cos B}+\dfrac{1+\cos B}{\sin B}=\dfrac{\sin^2 B+(1+\cos B)^2}{(1+\cos B)\sin B}[/tex]

Use [tex]\sin^2 B+\cos^2 B=1[/tex]:

[tex]\sin^2 B+1+2\cos B+\cos^2 B=2+2\cos B=2(1+\cos B)[/tex]

So

[tex]\dfrac{2(1+\cos B)}{(1+\cos B)\sin B}=\dfrac{2}{\sin B}=2\csc B[/tex]

This holds for [tex]\sin B\neq 0[/tex] and [tex]\cos B\neq -1[/tex]

[nycmath]

Hi, it's Bob...

[tex]\dfrac{\sin B}{1+\cos B}+\dfrac{1+\cos B}{\sin B}=\dfrac{\sin^2 B+(1+\cos B)^2}{(1+\cos B)\sin B}[/tex]

Use [tex]\sin^2 B+\cos^2 B=1[/tex]:

[tex]\sin^2 B+1+2\cos B+\cos^2 B=2+2\cos B=2(1+\cos B)[/tex]

So

[tex]\dfrac{2(1+\cos B)}{(1+\cos B)\sin B}=\dfrac{2}{\sin B}=2\csc B[/tex]

This holds for [tex]\sin B\neq 0[/tex] and [tex]\cos B\neq -1[/tex]

You hit the nail on the head.