How do you prove $\frac{csc B}{cos B} - \frac{cos B}{sin B} = tan B$ $csc B/cos B - cosB/sin B = tan B$ ?

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Answer 1 · 2015-11-26T03:04:35

It is explained below

$= \frac{csc B sin B - {cos}^{2} B}{sin B cos B}$ $=(csc B sin B- cos^2 B)/(sin B cos B)$

=(1-cos^2 B)/(sin B cos B)#

= $\frac{{sin}^{2} B}{sin B cos B}$ $(sin^2 B)/(sin B cos B)$
= $\frac{sin B}{cos B} = tan B$ $(sin B)/cos B= tan B$

Answer 2 · 2015-11-26T03:15:35

$\frac{csc B}{cos B} - \frac{cos B}{sin B} = tan B$ $cscB/cosB-cosB/sinB=tanB$

LS:
$\frac{csc B}{cos B} - \frac{cos B}{sin B}$ $cscB/cosB-cosB/sinB$

$= \frac{\frac{1}{sin B}}{cos B} - \frac{cos B}{sin B}$ $=(1/sinB)/cosB-cosB/sinB$

$= (\frac{1}{sin B} \div cos B) - \frac{cos B}{sin B}$ $=(1/sinB-:cosB)-cosB/sinB$

$= (\frac{1}{sin B} \cdot \frac{1}{cos B}) - \frac{cos B}{sin B}$ $=(1/sinB*1/cosB)-cosB/sinB$

$= \frac{1}{sin B cos B} - \frac{cos B}{sin B}$ $=1/(sinBcosB)-cosB/sinB$

$= \frac{1 - cos B (cos B)}{sin B cos B}$ $=(1-cosB(cosB))/(sinBcosB)$

$= \frac{1 - {cos}^{2} B}{sin B cos B}$ $=(1-cos^2B)/(sinBcosB)$

$= \frac{{sin}^{2} B}{sin B cos B}$ $=sin^2B/(sinBcosB)$

$= \frac{sin B sin B}{sin B cos B}$ $=(sinBsinB)/(sinBcosB)$

$=(color(red)cancelcolor(black)(sinB)sinB)/(color(red)cancelcolor(black)(sinB)cosB)$

$=tanB$

$:.$ , LS = RS.

How do you prove csc B/cos B - cosB/sin B = tan BcscBcosB−cosBsinB=tanBcsc B/cos B - cosB/sin B = tan B?