Given that tan x = 1, where 0o $\geq$ x 90o, evaluate $\frac{1 - \sin^2 x}{\cos x}$

Analyze the question: We're told tan x = 1 for x in the range 0° to 90° (the inequality is a typo, it should be 0° ≤ x ≤ 90°). We need to evaluate (1 − sin²x)/cos x.

Concept: The identity sin²x + cos²x = 1 gives 1 − sin²x = cos²x.

Solution:

1. Simplify the expression: (1 − sin²x)/cos x = cos²x / cos x = cos x

2. Find x from tan x = 1: tan x = 1 gives x = 45°

3. Evaluate cos 45°: cos 45° = √2/2 = 1/√2

So the expression equals √2/2.

Check with a triangle: tan x = 1 means opposite = adjacent = 1, so hypotenuse = √2. Then cos x = adjacent/hypotenuse = 1/√2 = √2/2. This agrees.

Answer: Option C, √2/2.

Tip: Whenever you see 1 − sin²x, think cos²x straight away, and 1 − cos²x gives sin²x. Also memorize the 45° values: sin 45° = cos 45° = √2/2 and tan 45° = 1. This saves time in the exam.