Question:
If $\tan \theta=\frac{12}{5}$, find the value of $\frac{1+\sin \theta}{1-\sin \theta}$.
Solution:
Given: $\tan \theta=\frac{12}{5}$
We have to find the value of the expression $\frac{1+\sin \theta}{1-\sin \theta}$.
$A C=\sqrt{A B^{2}+B C^{2}}$
$=\sqrt{12^{2}+5^{2}}$
$=13$
$\Rightarrow \sin \theta=\frac{12}{13}$
Therefore,
$\frac{1+\sin \theta}{1-\sin \theta}=\frac{1+\frac{12}{13}}{1-\frac{12}{13}}$
$=25$
Hence, the value of the given expression is 25.
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