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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