Question

Prove that : $(1 + \cot \theta - \csc \theta)(1 + \tan \theta + \sec \theta) = 2$

Hint:

Using the difference of squares identity $(a-b)(a+b) = a^2 - b^2$ is an extremely elegant and fast shortcut for simplifying products of trigonometric sums!

Answer 1

Step 1: Understanding the Question:
We are given a trigonometric identity to prove.
The expression on the Left-Hand Side (LHS) is a product of two binomial-like terms involving the ratios $\cot\theta$, $\csc\theta$, $\tan\theta$, and $\sec\theta$.
We need to show that this product simplifies to the constant value $2$ on the Right-Hand Side (RHS).

Step 2: Key Formula or Approach:
The most robust approach to simplify trigonometric identities is to convert all functions to their basic forms using sine ($\sin\theta$) and cosine ($\cos\theta$) relations:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta} \]
\[ \sec \theta = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{\sin \theta} \]
After conversion, we will perform algebraic simplification and use the fundamental identity:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]

Step 3: Detailed Explanation:

• Start by writing the Left-Hand Side (LHS) of the identity:
\[ \text{LHS} = (1 + \cot \theta - \csc \theta)(1 + \tan \theta + \sec \theta) \]

• Convert all trigonometric ratios to $\sin\theta$ and $\cos\theta$:
\[ \text{LHS} = \left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right) \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right) \]

• Find the common denominator for each of the two terms:
\[ \text{LHS} = \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right) \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \]

• Multiply the numerators and denominators:
\[ \text{LHS} = \frac{[(\sin \theta + \cos \theta) - 1][(\sin \theta + \cos \theta) + 1]}{\sin \theta \cos \theta} \]

• Apply the difference of squares identity $(a - b)(a + b) = a^2 - b^2$ where $a = \sin \theta + \cos \theta$ and $b = 1$:
\[ \text{LHS} = \frac{(\sin \theta + \cos \theta)^2 - (1)^2}{\sin \theta \cos A} \]

• Expand the squared term using $(x + y)^2 = x^2 + 2xy + y^2$:
\[ \text{LHS} = \frac{(\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta) - 1}{\sin \theta \cos \theta} \]

• Substitute the fundamental identity $\sin^2 \theta + \cos^2 \theta = 1$ into the numerator:
\[ \text{LHS} = \frac{(1 + 2\sin \theta \cos \theta) - 1}{\sin \theta \cos \theta} \]
\[ \text{LHS} = \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} \]

• Cancel the common factor $\sin\theta\cos\theta$ from the numerator and denominator:
\[ \text{LHS} = 2 = \text{RHS} \]


Step 4: Final Answer:
Since the Left-Hand Side simplifies exactly to $2$, the identity is proven.