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

Concept:
Use the half-angle identities:
Alternatively, multiply numerator and denominator inside the square root by \((1 + \cos\theta)\) to rationalize.

Step 1: Rationalize the expression.
Multiply numerator and denominator inside the square root by \((1 + \cos\theta)\):

Step 2: Simplify the denominator using the Pythagorean identity.
We know \(1 - \cos^2\theta = \sin^2\theta\). So:
Assuming \(0 < \theta < \pi\) (so \(\sin\theta > 0\) and \(1 + \cos\theta > 0\)):

Step 3: Split into standard trig functions.


Step 4: Verify with a specific angle.
Let \(\theta = 60^\circ\):

• LHS: \(\sqrt{\dfrac{1+\cos 60^\circ}{1-\cos 60^\circ}} = \sqrt{\dfrac{1+1/2}{1-1/2}} = \sqrt{\dfrac{3/2}{1/2}} = \sqrt{3}\).

• RHS: \(\csc 60^\circ + \cot 60^\circ = \dfrac{2}{\sqrt{3}} + \dfrac{1}{\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}\).

Step 5: Eliminate wrong options.

• \(\csc\theta = \dfrac{1}{\sin 60^\circ} = \dfrac{2}{\sqrt{3}} \approx 1.155 \neq \sqrt{3} \approx 1.732\).

• \(\cot\theta = \dfrac{1}{\sqrt{3}} \approx 0.577 \neq \sqrt{3}\).

• \(\csc\theta - \cot\theta = \dfrac{2}{\sqrt{3}} - \dfrac{1}{\sqrt{3}} = \dfrac{1}{\sqrt{3}} \neq \sqrt{3}\).