Question:
$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$ is equal to
(a) $\tan 90^{\circ}$
(b) 1
(c) $\sin 45^{\circ}$
(d) $\sin 0^{\circ}$
Solution:
We have to find the value of the following
$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$
So
$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}$
$=\frac{1-(1)^{2}}{1+(1)^{2}}$
$=\frac{0}{1}$
$=0$
We know that $\left[\begin{array}{r}\tan 45^{\circ}=1 \\ \sin 0^{\circ}=0\end{array}\right]$
$=\sin 0^{\circ}$
Hence the correct option is (d)