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