Question

The value of \[ \frac{1-\tan^2 45^\circ}{1+\tan^2 45^\circ} \] is

• A. 4
• B. 2
• C. 1
• D. 0

Hint:

Important standard values: \[ \sin45^\circ=\cos45^\circ=\frac{1}{\sqrt2}, \qquad \boxed{\tan45^\circ=1.} \]

Answer 1

The Correct Option is D

Concept: The standard trigonometric value is \[ \boxed{\tan 45^\circ=1.} \]

Step 1: Substitute the value of \(\tan45^\circ\).
Since, \[ \tan45^\circ=1, \] the given expression becomes \[ \frac{1-1^2}{1+1^2} = \frac{1-1}{1+1}. \]

Step 2: Simplify the expression.
\[ =\frac{0}{2}=0. \]

Step 3: Final conclusion.
Therefore, \[ \boxed{\frac{1-\tan^2 45^\circ}{1+\tan^2 45^\circ}=0.} \]