Math Problem Statement

prove (sintheta/(1+costheta))+(1+costheta)/(sintheta)=2csctheta

Solution

We are tasked with proving the identity:

sinθ1+cosθ+1+cosθsinθ=2cscθ\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \csc \theta

Step 1: Combine the terms on the left-hand side (LHS)

We start by simplifying the left-hand side (LHS) into a single fraction. To do this, we need a common denominator for the two fractions:

sinθ1+cosθ+1+cosθsinθ\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta}

The common denominator will be:

(1+cosθ)sinθ(1 + \cos \theta) \sin \theta

So, we rewrite each term with this common denominator:

sinθ1+cosθ=sin2θ(1+cosθ)sinθ\frac{\sin \theta}{1 + \cos \theta} = \frac{\sin^2 \theta}{(1 + \cos \theta) \sin \theta} 1+cosθsinθ=(1+cosθ)2(1+cosθ)sinθ\frac{1 + \cos \theta}{\sin \theta} = \frac{(1 + \cos \theta)^2}{(1 + \cos \theta) \sin \theta}

Now combine the two terms:

sin2θ+(1+cosθ)2(1+cosθ)sinθ\frac{\sin^2 \theta + (1 + \cos \theta)^2}{(1 + \cos \theta) \sin \theta}

Step 2: Simplify the numerator

Expand the square in the numerator:

(1+cosθ)2=1+2cosθ+cos2θ(1 + \cos \theta)^2 = 1 + 2 \cos \theta + \cos^2 \theta

So the numerator becomes:

sin2θ+1+2cosθ+cos2θ\sin^2 \theta + 1 + 2 \cos \theta + \cos^2 \theta

Now use the Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1:

1+1+2cosθ=2+2cosθ1 + 1 + 2 \cos \theta = 2 + 2 \cos \theta

Thus, the LHS is now:

2+2cosθ(1+cosθ)sinθ\frac{2 + 2 \cos \theta}{(1 + \cos \theta) \sin \theta}

Step 3: Factor the numerator

Factor out the common factor of 2:

2(1+cosθ)(1+cosθ)sinθ\frac{2(1 + \cos \theta)}{(1 + \cos \theta) \sin \theta}

Now, cancel 1+cosθ1 + \cos \theta from the numerator and denominator:

2sinθ\frac{2}{\sin \theta}

Step 4: Final simplification

We know that 1sinθ=cscθ\frac{1}{\sin \theta} = \csc \theta, so:

2sinθ=2cscθ\frac{2}{\sin \theta} = 2 \csc \theta

Thus, we have:

sinθ1+cosθ+1+cosθsinθ=2cscθ\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \csc \theta

This completes the proof.


Would you like to go over any specific details, or do you have further questions?

Related Questions:

  1. Can you prove the identity 1cosθsinθ=cotθ2\frac{1 - \cos \theta}{\sin \theta} = \cot \frac{\theta}{2}?
  2. How do trigonometric identities simplify expressions involving cscθ\csc \theta and secθ\sec \theta?
  3. What is the proof for sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1?
  4. How can we use the Pythagorean identity to simplify more complicated trigonometric expressions?
  5. What is the method for deriving sum and difference formulas for trigonometric functions?

Tip:

Remember that the Pythagorean identity, sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1, is very useful when simplifying trigonometric expressions or proving identities.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Identities
Simplification of Expressions

Formulas

sin^2(θ) + cos^2(θ) = 1
csc(θ) = 1/sin(θ)

Theorems

Pythagorean Identity

Suitable Grade Level

Grades 10-12