MCQ
$\frac{1-\tan^245^\circ}{1+\tan^245^\circ}$ is equal to :
- A$\tan90^\circ$
- B$1$
- C$\sin45^\circ$
- ✓$\sin0^\circ$
✓
Answer
Correct option: D.
$\sin0^\circ$
We have to find the value of the following
$\frac{1-\tan^245^\circ}{1+\tan^245^\circ}$
So,
$\frac{1-\tan^245^\circ}{1+\tan^245^\circ}$
$=\frac{1-(1)^2}{1+(1)^2}$
$=\frac{0}{1}$
$=0$
We know that $\begin{bmatrix}\tan45^\circ=1\\\sin0^\circ=0\end{bmatrix}$
$=\sin0^\circ$
Hence the correct option is $(d)$
$\frac{1-\tan^245^\circ}{1+\tan^245^\circ}$
So,
$\frac{1-\tan^245^\circ}{1+\tan^245^\circ}$
$=\frac{1-(1)^2}{1+(1)^2}$
$=\frac{0}{1}$
$=0$
We know that $\begin{bmatrix}\tan45^\circ=1\\\sin0^\circ=0\end{bmatrix}$
$=\sin0^\circ$
Hence the correct option is $(d)$
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