Question

Prove that : \(\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\tan A}{\sec A + 1}\).

Hint:

Converting all complex trigonometric functions like \(\tan A\), \(\sec A\), \(\cot A\) into basic \(\sin A\) and \(\cos A\) terms at the beginning is a very reliable strategy for solving any trigonometric identity.

Answer 1

Correct Answer: 4

Step 1: State what we need to prove.
We need to prove: \[ \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\tan A}{\sec A + 1} \]
Step 2: Work with the LHS - rationalise the expression.
\[ \text{LHS} = \sqrt{\frac{1 - \cos A}{1 + \cos A}} \times \sqrt{\frac{1 - \cos A}{1 - \cos A}} = \sqrt{\frac{(1 - \cos A)^2}{1 - \cos^2 A}} \]
Step 3: Use the identity $1 - \cos^2 A = \sin^2 A$.
\[ = \sqrt{\frac{(1 - \cos A)^2}{\sin^2 A}} = \frac{1 - \cos A}{\sin A} \quad (\text{taking positive square root}) \]
Step 4: Work with the RHS.
\[ \text{RHS} = \frac{\tan A}{\sec A + 1} = \frac{\frac{\sin A}{\cos A}}{\frac{1}{\cos A} + 1} = \frac{\frac{\sin A}{\cos A}}{\frac{1 + \cos A}{\cos A}} = \frac{\sin A}{1 + \cos A} \]
Step 5: Show LHS equals RHS.
Multiply numerator and denominator of LHS by $(1 + \cos A)$: \[ \frac{1 - \cos A}{\sin A} \times \frac{1 + \cos A}{1 + \cos A} = \frac{1 - \cos^2 A}{\sin A(1 + \cos A)} = \frac{\sin^2 A}{\sin A(1 + \cos A)} = \frac{\sin A}{1 + \cos A} = \text{RHS} \]
Step 6: Conclusion.
LHS = RHS. Hence the identity is proved.
\[ \boxed{\sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\tan A}{\sec A + 1} \text{ (proved)}} \]