Question

The expression $\frac{\tan x - 1 + \sec x}{\tan x - \sec x + 1}$ is equal to:}

• 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:

In rational trigonometric expressions with 1, $\tan$, and $\sec$, substituting $1 = \sec^2 - \tan^2$ usually leads to a cancellation.

Answer 1

The Correct Option is B

Step 1: Concept
Use the trigonometric identity $1 = \sec^2 x - \tan^2 x$ to simplify the numerator or denominator.

Step 2: Meaning
Rewrite the $1$ in the numerator: $(\tan x + \sec x) - (\sec^2 x - \tan^2 x)$.

Step 3: Analysis
Factorize: $(\tan x + \sec x) [1 - (\sec x - \tan x)] = (\tan x + \sec x)(1 - \sec x + \tan x)$. The denominator is $(\tan x - \sec x + 1)$, which cancels out the term in the numerator.

Step 4: Conclusion
The remaining term is $\sec x + \tan x = 1/\cos x + \sin x/\cos x = (1 + \sin x)/\cos x$.
Final Answer: (B)