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: Understanding the Question:
We are given a trigonometric identity to prove:
\[ \sqrt{\frac{1 - \cos A}{1 + \cos A}} = \frac{\tan A}{\sec A + 1} \]
We need to show that the Left-Hand Side (LHS) of this equation can be algebraically manipulated to match the Right-Hand Side (RHS).

Step 2: Key Formula or Approach:
1. Rationalizing the Radicand:
We will rationalize the denominator inside the square root of the LHS by multiplying the numerator and denominator by \((1 - \cos A)\).
2. Trigonometric Identities:
- \(\sin^2 A + \cos^2 A = 1 \implies 1 - \cos^2 A = \sin^2 A\)
- \(\tan A = \frac{\sin A}{\cos A}\)
- \(\sec A = \frac{1}{\cos A}\)

Step 3: Detailed Explanation:

• Let us begin by simplifying the Left-Hand Side (LHS):
\[ \text{LHS} = \sqrt{\frac{1 - \cos A}{1 + \cos A}} \] - Multiply the numerator and the denominator inside the square root by \((1 - \cos A)\):
\[ \text{LHS} = \sqrt{\frac{(1 - \cos A)(1 - \cos A)}{(1 + \cos A)(1 - \cos A)}} \] \[ \text{LHS} = \sqrt{\frac{(1 - \cos A)^2}{1 - \cos^2 A}} \]

• Use the identity \(1 - \cos^2 A = \sin^2 A\) in the denominator:
\[ \text{LHS} = \sqrt{\frac{(1 - \cos A)^2}{\sin^2 A}} \] - Take the square root of the numerator and the denominator:
\[ \text{LHS} = \frac{1 - \cos A}{\sin A} \] --- (Equation 1)

• Now, let us simplify the Right-Hand Side (RHS):
\[ \text{RHS} = \frac{\tan A}{\sec A + 1} \] - Express \(\tan A\) and \(\sec A\) in terms of sine and cosine:
\[ \text{RHS} = \frac{\frac{\sin A}{\cos A}}{\frac{1}{\cos A} + 1} \] - Simplify the denominator:
\[ \frac{1}{\cos A} + 1 = \frac{1 + \cos A}{\cos A} \] - Substitute this back into the RHS expression:
\[ \text{RHS} = \frac{\frac{\sin A}{\cos A}}{\frac{1 + \cos A}{\cos A}} \] - Cancel out the common denominator \(\cos A\):
\[ \text{RHS} = \frac{\sin A}{1 + \cos A} \] --- (Equation 2)

• To show that Equation 1 is equal to Equation 2, let us multiply the numerator and denominator of Equation 1 by \((1 + \cos A)\):
\[ \text{LHS} = \frac{(1 - \cos A)(1 + \cos A)}{\sin A(1 + \cos A)} \] \[ \text{LHS} = \frac{1 - \cos^2 A}{\sin A(1 + \cos A)} \] - Substitute \(1 - \cos^2 A = \sin^2 A\):
\[ \text{LHS} = \frac{\sin^2 A}{\sin A(1 + \cos A)} \] - Cancel one \(\sin A\) term from the numerator and denominator:
\[ \text{LHS} = \frac{\sin A}{1 + \cos A} \]

• Since both LHS and RHS simplify to \(\frac{\sin A}{1 + \cos A}\), we have:
\[ \text{LHS} = \text{RHS} \]


Step 4: Final Answer:
Hence, the identity is successfully proved.