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Simplify the following expression: 1−cosθsinθ+sinθ1−cosθ
2cscθ
2cosθ
2tanθ
2sinθ
2cscθ
By substituting θ = 90°, the expression becomes 1/1 + 0/1 = 1. Checking options: (A) 2sec90°=undefined, (B) 2cosec90°=2, (C) 2cos90°=0, (D) 2sin90°=2. However, the expression simplifies to 2cosecθ. Check θ = 30°: (1/2)/(1/2) + (1/2)/(1/2) = 1+1=2. 2cosec30° = 2(2) = 4. Let's re-simplify: (sin²θ + (1-cosθ)²)/(sinθ(1-cosθ)) = (sin²θ + 1 - 2cosθ + cos²θ)/(sinθ(1-cosθ)) = (2 - 2cosθ)/(sinθ(1-cosθ)) = 2(1-cosθ)/(sinθ(1-cosθ)) = 2/sinθ = 2cosecθ.
An algebraic expression in trigonometry: sinθ/(1-cosθ) + (1-cosθ)/sinθ
1−cosθsinθ+sinθ1−cosθ
By substituting θ = 90°, the expression becomes 1/1 + 0/1 = 1. Checking options: (A) 2sec90°=undefined, (B) 2cosec90°=2, (C) 2cos90°=0, (D) 2sin90°=2. However, the expression simplifies to 2cosecθ. Check θ = 30°: (1/2)/(1/2) + (1/2)/(1/2) = 1+1=2. 2cosec30° = 2(2) = 4. Let's re-simplify: (sin²θ + (1-cosθ)²)/(sinθ(1-cosθ)) = (sin²θ + 1 - 2cosθ + cos²θ)/(sinθ(1-cosθ)) = (2 - 2cosθ)/(sinθ(1-cosθ)) = 2(1-cosθ)/(sinθ(1-cosθ)) = 2/sinθ = 2cosecθ.
Many students try to rationalize the first term immediately, which is valid, but simply finding a common denominator is faster and less error-prone.
Finding a common denominator
Combine the two fractions by finding the common denominator, which is sinθ(1−cosθ).
sinθ(1−cosθ)sin2θ+(1−cosθ)2
Expanding and simplifying
Expand the numerator using (1−cosθ)2=1−2cosθ+cos2θ.
sinθ(1−cosθ)sin2θ+1−2cosθ+cos2θ
Applying trigonometric identities
Use the identity sin2θ+cos2θ=1 to simplify the numerator.
sinθ(1−cosθ)1+1−2cosθ=sinθ(1−cosθ)2−2cosθ
Final reduction
Factor out 2 from the numerator and cancel the common term (1−cosθ).
sinθ(1−cosθ)2(1−cosθ)=sinθ2=2cscθ
Option B is correct because the expression simplifies algebraically to 2cscθ.
This simplification technique is fundamental in solving more complex calculus limits involving trigonometric functions.