# Evaluating a Trigonometric Expression involving Periodicity

Evaluate:

$$\dfrac{\csc(90+\theta)+\cot(450+\theta)}{\csc(450-\theta)-\tan(180+\theta)}+\dfrac{\tan(180+\theta)+\sec(180-\theta)}{\tan(360-\theta)-\sec(-\theta)}$$

I simplified this into $$\dfrac{\sec(\theta)-\tan(\theta)}{\sec(\theta)-\tan(\theta)}+\dfrac{\tan(\theta)-\sec(\theta)}{-\tan(\theta)-\sec(\theta)}$$ $$1+\dfrac{\sec(\theta)-\tan(\theta)}{\sec(\theta)+\tan(\theta)}$$ $$1+\dfrac{\sec^2\theta+\tan^2\theta-2\sec\theta\tan\theta}{\sec^2\theta-\tan^2\theta}$$

Unfortunately I've got stuck here and cannot understand what to do. I'm not even sure if what I've done is correct; I was told that by the third step, the questions should be over.I would be truly grateful if somebody would kindly help me through solving this problem and also point out my errors. Many thanks in advance!

• @daryakhosrotash $\csc (90+\theta)=\sec \theta$. – Mark Viola Jul 21 '15 at 22:24
• @Makeadifference If the steps are correct, the last term can be simplified a bit to $\frac{2}{1+\sin \theta}$. Does that "Make a Difference?" ;-))) – Mark Viola Jul 21 '15 at 22:25
• @Dr.MV Your comment brought a huge smile to my face:)Unfortunately, I can't see how you got to that. Please could you elaborate a bit? Also, the final result (I am told) should be a constant. PS. I'm going to sleep now. It's 4 in the morning here in India! Goodnight:) – Make a Difference Jul 21 '15 at 22:32

$$1+\frac{sec \theta -tan \theta }{sec \theta +tan \theta }=\\1+\frac{sec \theta -tan \theta }{sec \theta +tan \theta }*\frac{cos \theta}{cos \theta}=\\1+\frac{\frac{1}{cos \theta} -\frac{sin \theta}{cos \theta} }{\frac{1}{cos \theta} +\frac{sin \theta}{cos \theta} }*\frac{cos \theta}{cos \theta} =\\1+\frac{1-sin \theta }{1+sin \theta}=\\\frac{1+sin \theta +1-sin \theta }{1+sin \theta}=\frac{2}{1+sin \theta}$$ second step :period $$f(x)=f(x+T)\\f(x)=\frac{2}{1+sin \theta}\\\frac{2}{1+sin \theta}=\frac{2}{1+sin (\theta+T)} \rightarrow sin(\theta)=sin(\theta+T ) \\T+\theta =\theta +2k \pi \rightarrow t=2k\pi \rightarrow T=2\pi$$