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Simplify the expression (1+cosθ)/sinθ * (cscθ - cotθ)

Mathematics

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Answer

B. 1\text{B. } 1

Step-by-step solution

10 steps

To simplify the expression 1+cos⁡θsin⁡θ(csc⁡θ−cot⁡θ)\frac{1+\cos\theta}{\sin\theta}(\csc\theta - \cot\theta), we will follow these steps:

  1. Rewrite csc⁡θ\csc\theta and cot⁡θ\cot\theta in terms of sin⁡θ\sin\theta and cos⁡θ\cos\theta.

    Recall the definitions:

    csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta} cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}
  2. Substitute these into the expression.

    1+cos⁡θsin⁡θ(1sin⁡θ−cos⁡θsin⁡θ)\frac{1+\cos\theta}{\sin\theta}\left(\frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta}\right)
  3. Simplify the term inside the parenthesis by finding a common denominator.

    1sin⁡θ−cos⁡θsin⁡θ=1−cos⁡θsin⁡θ\frac{1}{\sin\theta} - \frac{\cos\theta}{\sin\theta} = \frac{1-\cos\theta}{\sin\theta}
  4. Substitute this simplified term back into the main expression.

    1+cos⁡θsin⁡θ(1−cos⁡θsin⁡θ)\frac{1+\cos\theta}{\sin\theta}\left(\frac{1-\cos\theta}{\sin\theta}\right)
  5. Multiply the numerators and the denominators.

    (1+cos⁡θ)(1−cos⁡θ)sin⁡θ⋅sin⁡θ\frac{(1+\cos\theta)(1-\cos\theta)}{\sin\theta \cdot \sin\theta}
  6. Simplify the numerator using the difference of squares formula, (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2.

    Here, a=1a=1 and b=cos⁡θb=\cos\theta.

    (1+cos⁡θ)(1−cos⁡θ)=12−cos⁡2θ=1−cos⁡2θ(1+\cos\theta)(1-\cos\theta) = 1^2 - \cos^2\theta = 1 - \cos^2\theta
  7. Apply the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.

    From this identity, we know that 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta.
    So, the numerator becomes sin⁡2θ\sin^2\theta.

  8. Simplify the denominator.

    sin⁡θ⋅sin⁡θ=sin⁡2θ\sin\theta \cdot \sin\theta = \sin^2\theta
  9. Substitute the simplified numerator and denominator back into the expression.

    sin⁡2θsin⁡2θ\frac{\sin^2\theta}{\sin^2\theta}
  10. Simplify the fraction.

    Assuming sin⁡2θ≠0\sin^2\theta \neq 0, the expression simplifies to 1.

    The final answer is B. 1\text{B. } 1.

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