Question

Prove that : (sin A + sec A)\(^2\) + (cos A + cosec A)\(^2\) = (1 + sec A cosec A)\(^2\)

Hint:

An important algebraic-trigonometric shortcut to remember:
\[ \sec^2 A + \csc^2 A = \sec^2 A \csc^2 A \]
This identity converts a sum into a product and is extremely helpful in simplifying expressions containing sum of squared reciprocal functions.

Answer 1

Step 1: Understanding the Question:
The topic of this question is Trigonometric Identities.
We are asked to prove a complex trigonometric identity:
\[ (\sin A + \sec A)^2 + (\cos A + \csc A)^2 = (1 + \sec A \csc A)^2 \]

Step 2: Key Formula or Approach:
- Expand the left-hand side of the expression.
- Use standard reciprocal identities:
\[ \sec A = \frac{1}{\cos A}, \quad \csc A = \frac{1}{\sin A} \]
- Group the squared terms and use the Pythagorean identity \( \sin^2 A + \cos^2 A = 1 \) to simplify the expression.

Step 3: Detailed Explanation:
1. Write down the left-hand side (LHS) of the identity:
\[ \text{LHS} = (\sin A + \sec A)^2 + (\cos A + \csc A)^2 \]
2. Expand both squared terms using \( (x + y)^2 = x^2 + y^2 + 2xy \):
\[ \text{LHS} = (\sin^2 A + \sec^2 A + 2 \sin A \sec A) + (\cos^2 A + \csc^2 A + 2 \cos A \csc A) \]
3. Rearrange and group the terms for easier simplification:
\[ \text{LHS} = (\sin^2 A + \cos^2 A) + (\sec^2 A + \csc^2 A) + 2(\sin A \sec A + \cos A \csc A) \]
4. Substitute \( \sin^2 A + \cos^2 A = 1 \):
\[ \text{LHS} = 1 + (\sec^2 A + \csc^2 A) + 2\left(\sin A \cdot \frac{1}{\cos A} + \cos A \cdot \frac{1}{\sin A}\right) \]
\[ \text{LHS} = 1 + (\sec^2 A + \csc^2 A) + 2\left(\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}\right) \]
5. Simplify the fractional terms in the parenthesis:
\[ \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A} = \sec A \csc A \]
6. Simplify the term \( \sec^2 A + \csc^2 A \):
\[ \sec^2 A + \csc^2 A = \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A} = \frac{\sin^2 A + \cos^2 A}{\sin^2 A \cos^2 A} = \frac{1}{\sin^2 A \cos^2 A} = \sec^2 A \csc^2 A \]
7. Substitute these back into our LHS expression:
\[ \text{LHS} = 1 + \sec^2 A \csc^2 A + 2\sec A \csc A \]
8. Notice that this expression is a perfect square trinomial of the form \( a^2 + b^2 + 2ab = (a + b)^2 \), with \( a = 1 \) and \( b = \sec A \csc A \):
\[ \text{LHS} = (1 + \sec A \csc A)^2 \]
9. This is exactly equal to the right-hand side (RHS) of our equation.
This completes the proof.

Step 4: Final Answer:
Hence, it is proved that \((\sin A + \sec A)^2 + (\cos A + \csc A)^2 = (1 + \sec A \csc A)^2\).