Answer
$\dfrac{\csc \alpha+\sec \alpha }{\sin \alpha+\cos \alpha}=\cot \alpha +\tan \alpha=\dfrac{1 }{\sin \alpha+\cos \alpha}(proved)$
Work Step by Step
Prove the identity $\dfrac{\csc \alpha+\sec \alpha }{\sin \alpha+\cos \alpha}=\cot \alpha +\tan \alpha$
Consider the left side:$\dfrac{\csc \alpha+\sec \alpha }{\sin \alpha+\cos \alpha}=\dfrac{1/\sin \alpha+1/\cos \alpha }{\sin \alpha+\cos \alpha} =\dfrac{1 }{\sin \alpha+\cos \alpha}$
Now, consider the right side:$\cot \alpha +\tan \alpha=\dfrac{\cos \alpha }{\sin \alpha}+\dfrac{\sin \alpha }{\cos \alpha}=\dfrac{1 }{\sin \alpha+\cos \alpha}$
Thus, $\dfrac{\csc \alpha+\sec \alpha }{\sin \alpha+\cos \alpha}=\cot \alpha +\tan \alpha=\dfrac{1 }{\sin \alpha+\cos \alpha}$(proved)
