How do you prove that #costheta/(1-sintheta) + costheta/(1+sintheta) = 2 /costheta#?

Question

Zoe Austin · Accepted Answer

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#LHS : cos theta/(1-sin theta)+costheta/(1+sintheta)# #=(costheta(1+sintheta)+costheta(1-sintheta))/((1-sintheta)(1+sin theta))#->common denominator #=(costheta+sinthetacostheta+costheta-sinthetacostheta)/(1-sin^2theta)# #=(costheta+cancel(sinthetacostheta)+costheta-cancel(sinthetacostheta))/(1-sin^2theta)# #=(2costheta)/cos^2theta#->use the property #sin^2theta+cos^2theta=1# #=(2cancelcostheta)/cos^cancel2theta# #=2/costheta# #=RHS#

Josiah Anderson · Answer

undefined To prove $ \frac{\cos(\theta)}{1-\sin(\theta)} + \frac{\cos(\theta)}{1+\sin(\theta)} = \frac{2}{\cos(\theta)} $, follow these steps:

1. Start with the left-hand side of the equation:
$$ \frac{\cos(\theta)}{1-\sin(\theta)} + \frac{\cos(\theta)}{1+\sin(\theta)} $$

2. Find a common denominator for the fractions:
$$ \frac{\cos(\theta)(1+\sin(\theta))}{(1-\sin(\theta))(1+\sin(\theta))} + \frac{\cos(\theta)(1-\sin(\theta))}{(1-\sin(\theta))(1+\sin(\theta))} $$

3. Combine the fractions:
$$ \frac{\cos(\theta)(1+\sin(\theta)) + \cos(\theta)(1-\sin(\theta))}{(1-\sin(\theta))(1+\sin(\theta))} $$

4. Simplify the numerator:
$$ \cos(\theta) + \cos(\theta)\sin(\theta) + \cos(\theta) - \cos(\theta)\sin(\theta) $$

5. Combine like terms in the numerator:
$$ 2\cos(\theta) $$

6. Substitute the simplified numerator back into the equation:
$$ \frac{2\cos(\theta)}{(1-\sin(\theta))(1+\sin(\theta))} $$

7. Use the identity $ 1 - \sin^2(\theta) = \cos^2(\theta) $:
$$ \frac{2\cos(\theta)}{\cos^2(\theta)} $$

8. Simplify:
$$ \frac{2}{\cos(\theta)} $$

This is the right-hand side of the equation, thus proving the identity.