Question : If $\operatorname{tan} \theta=\frac{3}{4}$, then find the value of expression $\frac{1+\operatorname{sin} \theta}{1-\operatorname{sin} \theta}$.
Option 1: 4
Option 2: 3
Option 3: 8
Option 4: 5
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Correct Answer: 4
Solution :
Given: $\tan\theta=\frac{3}{4} =\frac{Perpendicular}{Base}$
Using Pythagoras' theorem,
$AC=\sqrt{AB^2+BC^2}$
⇒ $AC=\sqrt{3^2+4^2} =5$
$\sin\theta =\frac {\text {Perpendicular}}{\text {Hypotenuse} } = \frac{3}{5}$
Now,
$\frac{1+\sin\theta}{1-\sin\theta} = \frac{1+\frac{3}{5}}{1-\frac{3}{5}} =\frac{8}{2}=4$
Hence, the correct answer is 4.
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