Question

prove the following identity: \\( \frac { 1 - \cos ( 2 b ) } { \sin ( 2 b ) } = \frac { \tan ( b ) } { 1 } \\)
\\( \frac { 1 - \cos ( 2 b ) } { \sin ( 2 b ) } = \\)
\\( = \\)
\\( = \\)
\\( = \\)
\\( = \\)

Explanation:

Step1: Use double - angle formulas

We know that \(\cos(2b)=1 - 2\sin^{2}(b)\) and \(\sin(2b)=2\sin(b)\cos(b)\).
Substitute these into \(\frac{1-\cos(2b)}{\sin(2b)}\):

$$
LATEXBLOCK0
$$

Step2: Simplify the numerator

Simplify the numerator \(1-1 + 2\sin^{2}(b)=2\sin^{2}(b)\). So the expression becomes \(\frac{2\sin^{2}(b)}{2\sin(b)\cos(b)}\).

Step3: Cancel out common factors

Cancel out the common factor \(2\sin(b)\) (assuming \(\sin(b)
eq0\)). We get \(\frac{\sin(b)}{\cos(b)}\).

Step4: Use the definition of tangent

Since \(\tan(b)=\frac{\sin(b)}{\cos(b)}\), the left - hand side \(\frac{1-\cos(2b)}{\sin(2b)}=\tan(b)\).

Answer:

\(\frac{1-\cos(2b)}{\sin(2b)}=\frac{1-(1 - 2\sin^{2}(b))}{2\sin(b)\cos(b)}=\frac{2\sin^{2}(b)}{2\sin(b)\cos(b)}=\frac{\sin(b)}{\cos(b)}=\tan(b)\)