MCQ
$\frac{{1 - {{\tan }^2}45^\circ }}{{1 + {{\tan }^2}45^\circ }} = $
- A$\tan 90^{\circ}$
- B1
- C$\sin 45^{\circ}$
- D$0$
✓
Answer
$\begin{array}{l}\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}} \\
=\frac{1-(1)^2}{1+(1)^2} \\
=\frac{1-1}{1+1} \\
=\frac{0}{2} \\
=0\end{array}$
Hence, 0 is correct.
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