# Showing that if $x = \sec\theta + \tan\theta$, then $x + \frac{1}{x} = 2\sec\theta$

$x = \sec\theta + \tan\theta.$

Show that $x+\frac1x = 2\sec\theta$.

Thanks.

I used a few simple trig identities but get nowhere. I am confused about this question a lot. I do not see where the theta and x can be in one.

None are squared so I cannot used the identities I would think. The only thing I can think to do is

$x=\frac{1}{\sin\theta}+ \frac{\cos\theta}{\sin\theta}$ ...and then put them together.

• Your “only” idea is a good one. You can write $x = \frac{1+\cos\theta}{\sin\theta}$. – Matthew Leingang Sep 11 '17 at 15:41
• Notice that $$x + \frac{1}{x} = \sec\theta + \tan\theta + \frac{1}{\sec\theta + \tan\theta}$$ – N. F. Taussig Sep 11 '17 at 15:41
• @MatthewLeingang, except it should be $x+{1+\sin\theta\over\cos\theta}$. The OP got things a bit backwards. – Barry Cipra Sep 11 '17 at 15:44
• @BarryCipra: right you are. – Matthew Leingang Sep 11 '17 at 15:46
• I see - So the reciprocol of x reverses it and then I can manipulate it into the "show that"... – boi Shift Sep 11 '17 at 15:51

Your first problem is that $$\sec{\theta} = \frac{1}{\cos{\theta}}, \qquad \tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}},$$ so actually $$x = \frac{1}{\cos{\theta}} + \frac{\sin{\theta}}{\cos{\theta}} = \frac{1+\sin{\theta}}{\cos{\theta}}.$$ Then \begin{align} x + \frac{1}{x} &= \frac{1+\sin{\theta}}{\cos{\theta}} + \frac{\cos{\theta}}{1+\sin{\theta}} \\ &= \frac{(1+\sin{\theta})^2+\cos^2{\theta}}{\cos{\theta}(1+\sin{\theta})} \\ &= \frac{1 + 2\sin{\theta} + (\sin^2{\theta}+\cos^2{\theta})}{\cos{\theta}(1+\sin{\theta})} \\ &= \frac{2(1+\sin{\theta})}{\cos{\theta}(1+\sin{\theta})} = 2\sec{\theta}, \end{align} so in fact the identity is not true.
$$x=\frac1{\cos\theta}+\frac{\sin\theta}{\cos\theta}=\frac{1+\sin\theta}{\cos\theta}\quad\text{hence}\quad\frac1x=\dotsm$$ then reduce the fractions to the same denominator, and use some mid-school trigonometry.