Prove that \(1+\dfrac{\cot^2\alpha}{1+\csc\alpha}=\cosec\alpha\).
\(\displaystyle 1+\dfrac{\cot^2\alpha}{1+\csc\alpha}=\csc\alpha\).
Step 1: Recall the trigonometric identities.
Step 2: Start with the left-hand side (LHS).
\[ LHS = 1 + \dfrac{\cot^2\alpha}{1 + \csc\alpha} \]
Step 3: Replace \(\cot^2\alpha\) and \(\csc\alpha\) with their \(\sin\alpha\) and \(\cos\alpha\) forms.
\[ LHS = 1 + \dfrac{\dfrac{\cos^2\alpha}{\sin^2\alpha}}{1 + \dfrac{1}{\sin\alpha}} \]
Step 4: Simplify the denominator \(1 + \dfrac{1}{\sin\alpha}\).
\[ 1 + \dfrac{1}{\sin\alpha} = \dfrac{\sin\alpha + 1}{\sin\alpha} \]
Step 5: Put this back into the fraction.
\[ LHS = 1 + \dfrac{\cos^2\alpha}{\sin^2\alpha} \times \dfrac{\sin\alpha}{\sin\alpha + 1} \]
Step 6: Simplify the multiplication.
\[ LHS = 1 + \dfrac{\cos^2\alpha}{\sin\alpha(\sin\alpha + 1)} \]
Step 7: Write both terms of LHS with a common denominator.
\[ LHS = \dfrac{\sin\alpha(\sin\alpha + 1)}{\sin\alpha(\sin\alpha + 1)} + \dfrac{\cos^2\alpha}{\sin\alpha(\sin\alpha + 1)} \]
\[ LHS = \dfrac{\sin^2\alpha + \sin\alpha + \cos^2\alpha}{\sin\alpha(\sin\alpha + 1)} \]
Step 8: Use the Pythagoras identity \(\sin^2\alpha + \cos^2\alpha = 1\).
\[ LHS = \dfrac{1 + \sin\alpha}{\sin\alpha(\sin\alpha + 1)} \]
Step 9: Cancel \((1+\sin\alpha)\) from numerator and denominator.
\[ LHS = \dfrac{1}{\sin\alpha} \]
Step 10: Recall that \(\dfrac{1}{\sin\alpha} = \csc\alpha\).
Therefore, LHS = RHS = \(\csc\alpha\). ✔