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avatar+280 

I got this by canceling out with a conjugate,

 

 

 i

Veteran  Mar 21, 2017
 #1
avatar+26758 
+3

Remember that sec(t) = 1/cos(t) so multiply numerator and denominator by cos(t), and note that tan(t) = sin(t)/cos(t) and that 1 - cos(t)^2 = sin(t)^2:

 

\(\frac{\tan t}{\sec t-\cos t}\rightarrow\frac{\tan t\times \cos t}{1-\cos^2t}\rightarrow \frac{\sin t}{\sin^2 t}\rightarrow \frac{1}{\sin t}\rightarrow \csc t\)

.

Alan  Mar 21, 2017
 #2
avatar+19655 
+3

identity

 

 

Formula:

\(\begin{array}{|rcll|} \hline \sec(t) &=& \frac{1}{\cos(t)} \\ \tan(t) &=& \frac{\sin(t)}{\cos(t)} \\ \csc(t) &=& \frac{1}{\sin(t)} \\ \sin^2(t) &=& 1-\cos^2(t) \\ \hline \end{array}\)

 

\(\begin{array}{rcll} && \dfrac{ \tan(t) } {\sec(t)-\cos(t)} \\\\ &=& \dfrac{ \frac{\sin(t)}{\cos(t)} } {\frac{1}{\cos(t)}-\cos(t)} \\\\ &=& \dfrac{\sin(t)}{\cos(t)} \cdot \left( \frac{1} {\frac{1}{\cos(t)}-\cos(t)} \right) \\\\ &=& \dfrac{\sin(t)} {\frac{\cos(t)}{\cos(t)}-\cos(t)\cos(t)} \\\\ &=& \dfrac{\sin(t)} {1-\cos^2(t)} \\\\ &=& \dfrac{\sin(t)} {\sin^2(t)} \\\\ &=& \dfrac{1} {\sin(t)} \\\\ &=& \csc(t) \\ \end{array}\)

 

 

laugh

heureka  Mar 21, 2017

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