1. ## more basic trig

prove the following trig identity:

(1+sin theta/ cos theta) + (cos theta/ 1+sin theta) = 2/cos theta

Thank you!

2. Multiply the top & bottom of your second fraction by $\displaystyle 1-sin(\theta)$. Your denominator will then be $\displaystyle 1-sin^2(\theta)$, which equals $\displaystyle cos^2(\theta)$.

3. $\displaystyle \frac{1 + \sin x}{\cos x} + \frac{\cos x}{1 + \sin x} =$

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

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

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

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

$\displaystyle \frac{2 \cos x}{\cos^2 x} =$

$\displaystyle \frac{2}{\cos x}$

4. Originally Posted by ss103
prove the following trig identity:

(1+sin theta/ cos theta) + (cos theta/ 1+sin theta) = 2/cos theta

Thank you!
Working on the left hand side:

$\displaystyle \frac{1 + \sin x}{\cos x} + \frac{\cos x}{1 + \sin x} =$

$\displaystyle \frac{1+2\sin\theta+\sin^2\theta+\cos^2\theta}{\co s\theta(1+\sin\theta)}$

$\displaystyle \frac{1+2\sin\theta+1}{\cos\theta(1+\sin\theta)}$

$\displaystyle \frac{2+2\sin\theta}{\cos(1+\sin\theta)}$

$\displaystyle \frac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)}$

$\displaystyle \frac{2}{\cos\theta}$