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If $\tan \theta=x-\frac{1}{x}$, find $\sec \theta + \tan \theta$.

This was the question ask in my unit test.

My Efforts:

$\tan^2 \theta=(x-\frac{1}{x})^2$

$\tan^2 \theta= (\frac {x^2-1}{x})^2$

Now we can use identity $\sec^2 \theta= 1 + \tan^2 \theta$.

But i am not able to get the answer using this.

I don't know the correct answer but I had got $2x\ or\ -\frac{2}{x} $, which was given wrong.


Also please tell me if there is better way to do this.

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2 Answers 2

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As you have written $\sec^2 \theta = 1 + \tan^2 \theta$. From this you could get $$ \sec \theta + \tan \theta = \pm \sqrt{1+\tan^2 \theta} + \tan \theta = \pm \sqrt{1 + \left( x - \frac{1}{x} \right)^2} + x - \frac{1}{x}.$$

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    $\begingroup$ Nice answer. I was about to post an answer with $\pm$ sign. $\endgroup$
    – mike
    Aug 27, 2014 at 17:52
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You have the core of it. You are given that $ \tan \theta = x - \frac {1}{x} $, and you know that $ \sec^2 \theta = 1 + \tan^2 \theta $. Since $ \tan^2 \theta = x^2 + \frac {1}{x^2} - 2 $, you have $ \sec^2 \theta = x^2 + \frac {1}{x^2} - 1 $.

Therefore, $ \sec \theta = \sqrt {x^2 - \frac {1}{x^2} - 1} $ and $ \tan \theta = x - \frac {1}{x} $, so the answer is $$ \sec \theta + \tan \theta = x - \frac {1}{x} + \sqrt {x^2 + \frac {1}{x^2} - 1}. $$ I am not sure why you are saying the answer is $ 2x $ or $ - \frac {2}{x} $.

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  • $\begingroup$ I see your expresion for $\sec\theta+\tan\theta$ is wrong, I also wrote the - instead of +, did you copy? $\endgroup$
    – RE60K
    Aug 27, 2014 at 17:49
  • $\begingroup$ I have made silly mistake in solving, thanks. $\endgroup$ Aug 27, 2014 at 17:50
  • $\begingroup$ @Aditya LOL, definitely not. I didn't see your answer until I posted. Also, you're the one who had a mistake. Not me. You edited after I posted. $\endgroup$ Aug 27, 2014 at 17:50
  • $\begingroup$ it still is wrong $\endgroup$
    – RE60K
    Aug 27, 2014 at 17:51

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