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

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.

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  • $\begingroup$ Your “only” idea is a good one. You can write $x = \frac{1+\cos\theta}{\sin\theta}$. $\endgroup$ – Matthew Leingang Sep 11 '17 at 15:41
  • $\begingroup$ Notice that $$x + \frac{1}{x} = \sec\theta + \tan\theta + \frac{1}{\sec\theta + \tan\theta}$$ $\endgroup$ – N. F. Taussig 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. $\endgroup$ – Barry Cipra Sep 11 '17 at 15:44
  • $\begingroup$ @BarryCipra: right you are. $\endgroup$ – Matthew Leingang 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
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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.

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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.

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