Question #9385f

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Question #9385f

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Answer 1 · 2016-12-14T02:04:42

$\frac{csc B}{sin B} - \frac{cot B}{tan B} = 1$ $cscB/sinB - cotB/tanB = 1$

Use the identities $tan θ = \frac{sin θ}{cos θ}$ $tantheta = sin theta/costheta$ , $cot θ = \frac{1}{tan θ} = \frac{1}{\frac{sin θ}{cos θ}} = \frac{cos θ}{sin θ}$ $cottheta = 1/tantheta = 1/(sin theta/costheta) = costheta/sintheta$ and $csc θ = \frac{1}{sin θ}$ $csctheta = 1/sintheta$ .

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

$\Rightarrow \frac{1}{{sin}^{2} B} - \frac{{cos}^{2} B}{{sin}^{2} B} = 1$ $=>1/sin^2B - cos^2B/sin^2B = 1$

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

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

$\Rightarrow \frac{{sin}^{2} B}{{sin}^{2} B} = 1$ $=> sin^2B/sin^2B = 1$

$\Rightarrow 1 = 1$ $=> 1 = 1$

$L H S = R H S$ $LHS = RHS$

Identity proved!

Hopefully this helps!

Answer 2 · 2016-12-14T02:12:19

$\frac{csc B}{sin B} - \frac{cot B}{tan B} = 1$ $cscB/sinB-cotB/tanB=1$

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

$\frac{csc B}{\frac{1}{csc B}} - \frac{cot B}{tan B} - 1$ $frac(cscB)(1/cscB)-cotB/tanB-1$

$csc B \cdot \frac{csc B}{1} - \frac{cot B}{tan B} = 1$ $cscB*cscB/1-cotB/tanB=1$

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

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

${csc}^{2} B - \frac{cot B}{\frac{1}{cot B}} = 1$ $csc^2B-frac(cotB)(1/cotB)=1$

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

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

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

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

$1 = 1$ $1=1$

Answer 3 · 2016-12-14T03:31:57

See Below

Explanation:

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

Left Hand Side:

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

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

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

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

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

$:.=$ Right Hand Side

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Question #9385f

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