Question

\(\sqrt{\dfrac{1+\cos\theta}{1-\cos\theta}} =\)

• A. \(\csc\theta\)
• B. \(\cot\theta\)
• C. \(\csc\theta + \cot\theta\)
• D. \(\csc\theta - \cot\theta\)

Hint:

A standard trick: rationalize \(\sqrt{\dfrac{1+\cos\theta}{1-\cos\theta}}\) by multiplying inside by \(\dfrac{1+\cos\theta}{1+\cos\theta}\). This gives \(\dfrac{(1+\cos\theta)^2}{\sin^2\theta}\), and the square root simplifies to \(\dfrac{1+\cos\theta}{\sin\theta} = \csc\theta + \cot\theta\). Its reciprocal equals \(\csc\theta - \cot\theta\).

Answer 1

The Correct Option is C

Step 1: Plan the approach.
We want to simplify $\sqrt{\dfrac{1+\cos\theta}{1-\cos\theta}}$. A clean trick is to multiply top and bottom inside the root by $1 + \cos\theta$, which creates a perfect square.

Step 2: Multiply by the conjugate.
\[ \frac{1+\cos\theta}{1-\cos\theta}\times \frac{1+\cos\theta}{1+\cos\theta} = \frac{(1+\cos\theta)^2}{1-\cos^2\theta} \]

Step 3: Use the Pythagorean identity.
Since $1 - \cos^2\theta = \sin^2\theta$, the expression becomes $\dfrac{(1+\cos\theta)^2}{\sin^2\theta}$.

Step 4: Take the square root.
The square root of a square gives \[ \frac{1+\cos\theta}{\sin\theta} \] for the usual range where the quantities are positive.

Step 5: Split the fraction.
\[ \frac{1}{\sin\theta} + \frac{\cos\theta}{\sin\theta} = \csc\theta + \cot\theta \]

Step 6: State the result.
So the expression simplifies to $\csc\theta + \cot\theta$. Therefore \[ \boxed{\csc\theta + \cot\theta} \]