Question
$\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=?$
✓
Answer
$\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\frac{1-(1)^2}{1+(1)^2}\ldots\ldots \left[\because \tan 45^{\circ}=1\right]$
$=\frac{1-1}{1+1}$
$=\frac{0}{2}$
$=0$
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