Can someone please prove the following for me with detailed steps so I can understand?:
cot P + tan P = (cosec P)(sec P)
Any help would be greatly appreciated!
Thanks in advance!
$\displaystyle \cot{\theta} + \tan{\theta} = \frac{1}{\tan{\theta}} + \tan{\theta} = \frac{1+ \tan^2{\theta}}{\tan{\theta}} = \frac{1 + \left( \frac{\sin{\theta}}{\cos{\theta}} \right) ^2 }{ \frac{\sin{\theta}}{ \cos{\theta}}}$
I think this is more than enough help, you should finish the problem for yourself.
Hope this helped,
pomp.
Here's another way to do it:
$\displaystyle \cot P + \tan P$
$\displaystyle \begin{aligned}
&= \frac{\cos P}{\sin P} + \frac{\sin P}{\cos P} \\
&= \frac{\cos^2 P}{\sin P \cos P} + \frac{\sin^2 P}{\sin P \cos P} \\
&...
\end{aligned}$
I'll let you figure out the rest.
01