Question

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

Hint:

Rearranging terms is a powerful algebraic trick to simplify trigonometric proofs. It allows you to use conjugate identities like \((\sec x - \tan x)(\sec x + \tan x) = 1\) directly in a single step!

Answer 1

Step 1: Rationalise the first fraction on the LHS.
Multiply $\frac{1}{\sec x - \tan x}$ by $\frac{\sec x + \tan x}{\sec x + \tan x}$: since $\sec^2 x - \tan^2 x = 1$, this becomes $\sec x + \tan x$.
Step 2: Simplify the LHS using this result.
\[ \text{LHS} = (\sec x + \tan x) - \frac{1}{\cos x} = (\sec x + \tan x) - \sec x = \tan x \]
Step 3: Rationalise the fraction on the RHS the same way.
Multiply $\frac{1}{\sec x + \tan x}$ by $\frac{\sec x - \tan x}{\sec x - \tan x}$ to get $\sec x - \tan x$.
Step 4: Simplify the RHS and compare.
\[ \text{RHS} = \frac{1}{\cos x} - (\sec x - \tan x) = \sec x - \sec x + \tan x = \tan x \]
Since both sides simplify to $\tan x$, the identity is proved.
\[ \boxed{\tan x} \]