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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!

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  • $\begingroup$ @daryakhosrotash $\csc (90+\theta)=\sec \theta$. $\endgroup$
    – Mark Viola
    Jul 21, 2015 at 22:24
  • $\begingroup$ @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?" ;-))) $\endgroup$
    – Mark Viola
    Jul 21, 2015 at 22:25
  • $\begingroup$ @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:) $\endgroup$
    – Make a Difference
    Jul 21, 2015 at 22:32

1 Answer 1

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

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