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

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


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.

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


Here is a detailed answer.
Let's rock! $$\require{cancel} \begin{align} \sec\theta+\tan\theta+\frac{1}{\sec\theta+\tan\theta}&=\frac{\sec\theta\left(\sec\theta+\tan\theta\right)+\tan\theta\left(\sec\theta+\tan\theta\right)+1}{\sec\theta+\tan\theta}\\ &=\frac{\left(\sec\theta+\tan\theta\right)\left(\sec\theta+\tan\theta\right)+\color{red}1}{\sec\theta+\tan\theta}\\ &=\frac{\left(\sec\theta+\tan\theta\right)\left(\sec\theta+\tan\theta\right)+\color{red}{\sec^2\theta-\tan^2\theta}}{\sec\theta+\tan\theta}\\ &=\frac{\sec^2\theta+2\sec\theta\tan\theta\cancel{+\tan^2\theta}+\sec^2\theta\cancel{-\tan^2\theta}}{\sec\theta+\tan\theta}\\ &=\frac{2\sec^2\theta+2\sec\theta\tan\theta}{\sec\theta+\tan\theta}\\ &=\frac{2\sec\theta\cancel{\left(\sec\theta+\tan\theta\right)}}{\cancel{\sec\theta+\tan\theta}}\\ &=2\sec\theta \end{align} $$ We substituted $\color{red}1$ with $\color{red}{\sec^2\theta-\tan^2\theta}$.
I hope this helps.


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