+0  
 
+1
166
2
avatar+253 

I got this by canceling out with a conjugate,

 

 

 i

Veteran  Mar 21, 2017
Sort: 

2+0 Answers

 #1
avatar+26250 
+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+18626 
+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

7 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details