Answer:
proof given below
Step-by-step explanation:
Combine the fractions by making the denominators the same:
[tex]\dfrac{\sin \theta}{1+ \cos \theta}+\dfrac{1+ \cos \theta}{ \sin \theta}[/tex]
[tex]=\dfrac{\sin \theta}{1+ \cos \theta} \cdot \dfrac{\sin \theta}{\sin \theta}+\dfrac{1+ \cos \theta}{ \sin \theta} \cdot \dfrac{1+ \cos \theta}{1+ \cos \theta}[/tex]
[tex]=\dfrac{\sin \theta(\sin \theta)}{\sin \theta(1+ \cos \theta)}+\dfrac{(1+ \cos \theta)(1+ \cos \theta)}{\sin \theta(1+ \cos \theta)}[/tex]
[tex]= \dfrac{\sin^2 \theta}{\sin \theta(1+ \cos \theta)} +\dfrac{1+2 \cos \theta + \cos^2 \theta}{\sin \theta(1+ \cos \theta)}[/tex]
[tex]= \dfrac{\sin^2 \theta+ \cos^2 \theta+1+2 \cos \theta }{\sin \theta(1+ \cos \theta)}[/tex]
Use the trigonometric identity [tex]\sin^2 \theta + \cos^2 \theta=1[/tex] :
[tex]= \dfrac{1+1+2 \cos \theta}{\sin \theta(1+ \cos \theta)}[/tex]
[tex]= \dfrac{2+2 \cos \theta}{\sin \theta(1+ \cos \theta)}[/tex]
Factor out 2 from the numerator:
[tex]= \dfrac{2(1+ \cos \theta)}{\sin \theta(1+ \cos \theta)}[/tex]
Cancel the common factor:
[tex]= \dfrac{2}{\sin \theta}[/tex]
[tex]\textsf{Use the identity} \quad\csc \theta=\dfrac{1}{\sin \theta}:[/tex]
[tex]=2 \csc \theta[/tex]
Hence proved.
Learn more about trigonometric identities here:
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