0
$\begingroup$

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

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.

$\endgroup$
5
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

Hint:

$$x=\frac1{\cos\theta}+\frac{\sin\theta}{\cos\theta}=\frac{1+\sin\theta}{\cos\theta}\quad\text{hence}\quad\frac1x=\dotsm$$ then reduce the fractions to the same denominator, and use some mid-school trigonometry.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.