Answer
Please refer to the step-by-step part for the proof.
Work Step by Step
RECALL:
(1) $\csc{\theta}=\dfrac{1}{\sin{\theta}}$
(2) $\cot{\theta} = \dfrac{\cos{\theta}}{\sin{\theta}}$
Work on the left hand side of the equation to obtain:
$\require{cancel}
\sin{\theta}(\csc{\theta} + \cot{\theta})
\\=\sin{\theta}\left(\dfrac{1}{\sin{\theta}} + \dfrac{\cos{\theta}}{\sin{\theta}}\right)
\\=\sin{\theta}\cdot \dfrac{1}{\sin{\theta}} + \sin{\theta} \cdot
\dfrac{\cos{\theta}}{\sin{\theta}}
\\=\cancel{\sin{\theta}}\cdot \dfrac{1}{\cancel{\sin{\theta}}} + \cancel{\sin{\theta}} \cdot
\dfrac{\cos{\theta}}{\cancel{\sin{\theta}}}
\\=1+\cos{\theta}$
The two sides of the given equation was shown to be equal, therefore
$\sin{\theta}(\csc{\theta} + \cot{\theta}) = 1+\cos{\theta}$.