# Find $\sec \theta + \tan \theta$.

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.

As you have written $\sec^2 \theta = 1 + \tan^2 \theta$. From this you could get $$\sec \theta + \tan \theta = \pm \sqrt{1+\tan^2 \theta} + \tan \theta = \pm \sqrt{1 + \left( x - \frac{1}{x} \right)^2} + x - \frac{1}{x}.$$
• Nice answer. I was about to post an answer with $\pm$ sign. – mike Aug 27 '14 at 17:52
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 $- \frac {2}{x}$.
• I see your expresion for $\sec\theta+\tan\theta$ is wrong, I also wrote the - instead of +, did you copy? – RE60K Aug 27 '14 at 17:49