Question

\[ \frac{\tan x-1+\sec x}{\tan x-\sec x+1} = \]

• A. \( \frac{1-\sin x}{\cos x} \)
• B. \( \frac{1+\sin x}{\cos x} \)
• C. \( \frac{1+\cos x}{\sin x} \)
• D. \( \frac{1-\cos x}{\sin x} \)

Hint:

When \(\tan x\) and \(\sec x\) are present together, convert them into \(\sin x\) and \(\cos x\) to simplify easily.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This problem requires simplifying a trigonometric expression containing tangents and secants into a simpler form involving sines and cosines.
Step 2: Key Formula or Approach:
A clever trick for expressions of the form \( \frac{\tan \theta + \sec \theta - 1}{\tan \theta - \sec \theta + 1} \) is to substitute \( 1 = \sec^2 \theta - \tan^2 \theta \) only in the numerator.
Step 3: Detailed Explanation:

Rewrite the numerator as \( (\tan x + \sec x) - 1 \).

Substitute \( 1 = \sec^2 x - \tan^2 x \):
\[ \text{Num} = (\tan x + \sec x) - (\sec^2 x - \tan^2 x) \]

Factorize the difference of squares:
\[ \text{Num} = (\sec x + \tan x) - [(\sec x - \tan x)(\sec x + \tan x)] \]

Take \( (\sec x + \tan x) \) as a common factor:
\[ \text{Num} = (\sec x + \tan x) [1 - (\sec x - \tan x)] \]
\[ \text{Num} = (\sec x + \tan x) (1 - \sec x + \tan x) \]

Now look at the denominator: \( \tan x - \sec x + 1 \). This matches the second factor in our numerator exactly!

Divide numerator by denominator:
\[ \frac{(\sec x + \tan x) (1 - \sec x + \tan x)}{1 - \sec x + \tan x} = \sec x + \tan x \]

Convert to sine and cosine:
\[ \sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} \]

Step 4: Final Answer:
The expression simplifies to \( \frac{1 + \sin x}{\cos x} \).