Question

how would you prove this trigonomic identity? Please show steps.

$\frac{ \cos( \alpha ) }{1 + \sin( \alpha ) } + \frac{ \cos( \alpha ) }{1 - \sin( \alpha ) } = 2 \sec( \alpha )$
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Answer 1

Start by combining the fractions:

$\dfrac{\cos\alpha}{1+\sin\alpha}\cdot\dfrac{1-\sin\alpha}{1-\sin\alpha}+\dfrac{\cos\alpha}{1-\sin\alpha}\cdot\dfrac{1+\sin\alpha}{1+\sin\alpha}$

$\dfrac{\cos\alpha(1-\sin\alpha)+\cos\alpha(1+\sin\alpha)}{(1+\sin\alpha)(1-\sin\alpha)}$

$\dfrac{2\cos\alpha}{1-\sin^2\alpha}$

Recall the Pythagorean identity:

$\dfrac{2\cos\alpha}{\cos^2\alpha}$

Then cancel a factor of $\cos\alpha$ and use the definition of secant:

$\dfrac2{\cos\alpha}=\boxed{2\sec\alpha}$