8. Introduction to Trigonometry
easy

$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$

A

$\tan 90^{\circ}$

B

$1$

C

$0$

D

$\sin 45^{\circ}$

Solution

$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$

$=\frac{1-(1)^{2}}{1+(1)^{2}}=\frac{1-1}{1+1}=\frac{0}{2}=0$

Standard 10
Mathematics

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