Simple modification of complex expression

Question

I need to simplify $$ {\frac{1+i\tan\alpha}{1-i\tan\alpha}} $$ I solved it: $$ {\frac{1+i\tan\alpha}{1-i\tan\alpha}} = {\frac{(1+i\tan\alpha)(1+i\tan\alpha)}{(1-i\tan\alpha)(1+i\tan\alpha)}} = {\frac{1-\tan^2\alpha+i(\tan\alpha+\tan\alpha)}{(1+i\tan^2\alpha)(1+i\tan\alpha)}} = {\frac{1-\tan^2\alpha+i(\tan\alpha+\tan\alpha)}{\sec^2\alpha}} = {\frac{1-\tan^2\alpha}{\sec^2\alpha}} + i{\frac{2\tan\alpha}{\sec^2\alpha}} $$ The answer in the book is $$ \cos2\alpha+i\sin2\alpha (\alpha\ne{\frac{\pi}{2}}+\pi k) $$ What's wrong here?

Answer 1

What you did is fine. Now$$\frac{1-\tan^2\alpha}{\sec^2\alpha}=\cos^2\alpha-\sin^2\alpha=\cos(2\alpha)$$and$$2\frac{\tan\alpha}{\sec^2\alpha}=2\sin\alpha\cos\alpha=\sin(2\alpha).$$

Answer 2

Since $$\frac{1-\tan^2(x)}{\sec^2(x)}=\cos(2x)$$

Answer 3

An easy and direct way is to multiply both the numerator and the denominator by $ \cos \alpha$. This can be done as long as $\cos\alpha\neq 0$, i.e., $\alpha\neq \frac{\pi}{2} + \pi k$.

We obtain $$ \frac{1+i \tan\alpha}{1-i \tan\alpha} = \frac{\cos\alpha +i \sin\alpha}{\cos\alpha -i \sin\alpha} = \frac{e^{i\alpha}}{e^{-i\alpha}} = e^{2i\alpha} = \cos(2\alpha)+i \sin(2\alpha)\,.$$

Answer 4

You made some small typos which overall do not affect the final solution. I will write out a complete one to make everything more clear and convincible. Starting with the same as you we get \begin{align} &\color{red}{\frac{1+i\tan\alpha}{1-i\tan\alpha}}=\frac{1+i\tan\alpha}{1-i\tan\alpha}\cdot\frac{1+i\tan\alpha}{1+i\tan\alpha}=\frac{(1+i\tan\alpha)^2}{1-i^2\tan^2\alpha}=\frac{1-\tan^2\alpha+2i\tan\alpha}{1+\tan^2\alpha}\\ &=\frac{1-\tan^2\alpha}{1+\tan^2\alpha}+i\frac{2\tan\alpha}{1+\tan^2\alpha}=\frac{1-\frac{\sin^2\alpha}{\cos^2\alpha}}{1+\frac{\sin^2\alpha}{\cos^2\alpha}}+i\frac{2\frac{\sin\alpha}{\cos\alpha}}{1+\frac{\sin^2\alpha}{\cos^2\alpha}}=\frac{\frac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}{\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}+i\frac{2\frac{\sin\alpha\cos\alpha}{\cos^2\alpha}}{\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}\\ &=\frac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha}+i\frac{2\sin\alpha\cos\alpha}{\cos^2\alpha+\sin^2\alpha}=\frac{\cos(2\alpha)}1+i\frac{\sin(2\alpha)}1=\color{red}{\cos(2\alpha)+i\sin(2\alpha)} \end{align}

The restriction for $\alpha$ is natural due the fact that for $\alpha=\frac\pi2+k\pi,~k\in\mathbb Z$ the tangent function has a pole of first order.