# 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 $\csc (90+\theta)=\sec \theta$. Jul 21, 2015 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?" ;-))) Jul 21, 2015 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:) Jul 21, 2015 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 $$f(x)=f(x+T)\\f(x)=\frac{2}{1+sin \theta}\\\frac{2}{1+sin \theta}=\frac{2}{1+sin (\theta+T)} \rightarrow sin(\theta)=sin(\theta+T ) \\T+\theta =\theta +2k \pi \rightarrow t=2k\pi \rightarrow T=2\pi$$