Proving $\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$

Question

Prove this trigonometric identity:

$$\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$$

I've simplified it until

$$\frac{2\cos^2\theta}{1-\sin\theta}$$

but couldn't get $2+2\tan\theta$ from it.

Answer 1

We have that

$$\frac{1+\cos\theta}{\sec\theta-\tan\theta}+\frac{\cos\theta-1}{\sec\theta+\tan\theta}=\frac{\cos\theta(1+\cos\theta)}{1-\sin\theta}+\frac{\cos\theta(\cos\theta-1)}{1+\sin\theta}=$$

$$=\frac{\cos\theta(1+\cos\theta)(1+\sin\theta)+\cos\theta(\cos\theta-1)(1-\sin\theta)}{1-\sin^2\theta}=$$

$$=\frac{(1+\cos\theta)(1+\sin\theta)+(\cos\theta-1)(1-\sin\theta)}{\cos\theta}=$$

$$=\frac{2\cos \theta+2\sin \theta}{\cos\theta}=2+2\tan\theta$$

Answer 2

First, let's make a common denominator of $\frac{1+\cos\theta}{\sec\theta-\tan\theta}+\frac{\cos\theta-1}{\sec\theta+\tan\theta}$.
Ready, set, go! $$\require{cancel}\begin{aligned}\color{blue}{\frac{1+\cos\theta}{\sec\theta-\tan\theta}}+\color{red}{\frac{\cos\theta-1}{\sec\theta+\tan\theta}}&=\frac{\color{blue}{\left(1+\cos\theta\right)\left(\sec\theta+\tan\theta\right)}+\color{red}{\left(\cos\theta-1\right)\left(\sec\theta-\tan\theta\right)}}{\color{blue}{\left(\sec\theta+\tan\theta\right)}\color{red}{\left(\sec\theta-\tan\theta\right)}}\\&=\frac{\color{blue}{\cancel{\sec x}+\tan x+\cos{\theta}\sec{\theta}\cancel{+\cos{\theta}}\tan{\theta}}+\color{red}{\cos{\theta}\sec{\theta}\cancel{-\cos{\theta}\tan{\theta}}\cancel{-\sec x}+\tan x}}{\color{blue}{\left(\sec x+\tan x\right)}\color{red}{\left(\sec x-\tan x\right)}}\\&=\frac{2\tan x+2\cos{\theta}\sec{\theta}}{\sec^{2}x-\tan^{2}x}&&\text{The denominator is equal to 1.}\\&=\frac{2\tan x+2\cos{\theta}\sec{\theta}}{1}\\&=2\tan x+2\cos{\theta}\sec{\theta}\\&=2\tan\theta+2\cancel{\cos\theta}\cdot\frac{1}{\cancel{\cos\theta}}\\&=2+2\tan\theta\end{aligned}$$ I hope this helps.