# Question #9385f

Dec 14, 2016

$\csc \frac{B}{\sin} B - \cot \frac{B}{\tan} B = 1$

Use the identities $\tan \theta = \sin \frac{\theta}{\cos} \theta$, $\cot \theta = \frac{1}{\tan} \theta = \frac{1}{\sin \frac{\theta}{\cos} \theta} = \cos \frac{\theta}{\sin} \theta$ and $\csc \theta = \frac{1}{\sin} \theta$.

$\implies \frac{\frac{1}{\sin} B}{\sin} B - \frac{\cos \frac{B}{\sin} B}{\sin \frac{B}{\cos} B} = 1$

$\implies \frac{1}{\sin} ^ 2 B - {\cos}^{2} \frac{B}{\sin} ^ 2 B = 1$

$\implies \frac{1 - {\cos}^{2} B}{\sin} ^ 2 B = 1$

Use the identity ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$.

$\implies {\sin}^{2} \frac{B}{\sin} ^ 2 B = 1$

$\implies 1 = 1$

$L H S = R H S$

Identity proved!

Hopefully this helps!

Dec 14, 2016

$\csc \frac{B}{\sin} B - \cot \frac{B}{\tan} B = 1$

Use the identity $\sin B = \frac{1}{\csc} B$

$\frac{\csc B}{\frac{1}{\csc} B} - \cot \frac{B}{\tan} B - 1$

$\csc B \cdot \csc \frac{B}{1} - \cot \frac{B}{\tan} B = 1$

${\csc}^{2} B - \cot \frac{B}{\tan} B = 1$

Use the identity $\tan B = \frac{1}{\cot} B$

${\csc}^{2} B - \frac{\cot B}{\frac{1}{\cot} B} = 1$

${\csc}^{2} B - \cot B \cdot \cot \frac{B}{1} = 1$

${\csc}^{2} B - {\cot}^{2} B = 1$

Use the Pythagorean identity $1 + {\cot}^{2} B = {\csc}^{2} B$

$1 + {\cot}^{2} B - {\cot}^{2} B = 1$

$1 = 1$

Dec 14, 2016

See Below

#### Explanation:

$\csc \frac{B}{\sin} B - \cot \frac{B}{\tan} B = 1$

Left Hand Side:

$= \csc \frac{B}{\sin} B - \cot \frac{B}{\tan} B$

$= \csc B \cdot \frac{1}{\sin} B - \cot B \cdot \frac{1}{\tan} B$

$= \csc B \csc B - \cot B \cot B$

$= {\csc}^{2} B - {\cot}^{2} B$

$= 1$ -->Use the identity ${\cot}^{2} B + 1 = {\csc}^{2} B$ and subtract ${\cot}^{2} B$ from both sides to isolate 1.

$\therefore =$ Right Hand Side