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