Prove that \(\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=2\csc\theta\).
\(\displaystyle \dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}=2\csc\theta\).
Step 1: We start with the left-hand side (LHS):
\(\dfrac{\sin\theta}{1+\cos\theta}+\dfrac{1+\cos\theta}{\sin\theta}\)
Step 2: To add these fractions, we take the common denominator, which is \(\sin\theta(1+\cos\theta)\).
Step 3: Write the fractions with this denominator:
\(\dfrac{\sin^2\theta + (1+\cos\theta)^2}{\sin\theta(1+\cos\theta)}\)
Step 4: Expand the numerator:
\(\sin^2\theta + (1+\cos\theta)^2 = \sin^2\theta + (1 + 2\cos\theta + \cos^2\theta)\)
Step 5: Recall the Pythagoras identity: \(\sin^2\theta + \cos^2\theta = 1\).
So, \(\sin^2\theta + \cos^2\theta = 1\).
Step 6: Replace in the numerator:
\(1 + 1 + 2\cos\theta = 2 + 2\cos\theta\)
Step 7: Factorize:
\(2 + 2\cos\theta = 2(1+\cos\theta)\)
Step 8: Now the fraction becomes:
\(\dfrac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)}\)
Step 9: Cancel \((1+\cos\theta)\) from numerator and denominator:
\(\dfrac{2}{\sin\theta}\)
Step 10: Recall that \(\csc\theta = \dfrac{1}{\sin\theta}\).
Final Step: Therefore,
\(\dfrac{2}{\sin\theta} = 2\csc\theta\).
Hence proved.