Question

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

Hint:

In proofs like this, if standard simplification is messy, try shifting terms to make the equation symmetric. Proving \(A - B = C - D\) is the same as proving \(A + D = B + C\).

Answer 1

Step 1: Understanding the Concept:
We use trigonometric identities, specifically the relationship \(\sec^2 x - \tan^2 x = 1\), to rationalize denominators and simplify terms.
Step 2: Key Formula or Approach:
Rearrange the equation to group similar terms:
\[ \frac{1}{\sec x - \tan x} + \frac{1}{\sec x + \tan x} = \frac{1}{\cos x} + \frac{1}{\cos x} \]
We will prove this equality.
Step 3: Detailed Explanation:
Take the Left Hand Side (LHS) of the rearranged equation:
\[ \text{LHS} = \frac{1}{\sec x - \tan x} + \frac{1}{\sec x + \tan x} \]
Taking LCM:
\[ \text{LHS} = \frac{(\sec x + \tan x) + (\sec x - \tan x)}{(\sec x - \tan x)(\sec x + \tan x)} \]
\[ \text{LHS} = \frac{2 \sec x}{\sec^2 x - \tan^2 x} \]
Since \(\sec^2 x - \tan^2 x = 1\):
\[ \text{LHS} = 2 \sec x = \frac{2}{\cos x} \]
Now, evaluate the Right Hand Side (RHS) of the rearranged equation:
\[ \text{RHS} = \frac{1}{\cos x} + \frac{1}{\cos x} = \frac{2}{\cos x} \]
LHS = RHS. Hence, the original equation is proved.
Step 4: Final Answer:
Hence proved.