Question : In $\triangle ABC$, right angled at B, if $\tan A=\frac{1}{2}$, then the value of $\frac{\sin A(\cos C+\cos A)}{\cos C(\sin C-\sin A)}$ is:
Option 1: $2$
Option 2: $1$
Option 3: $3$
Option 4: $2\sqrt5$
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Correct Answer: $3$
Solution :
$\tan A=\frac{1}{2}$
$⇒\frac{BC}{AB}=\frac{1}{2}$
Let $BC=x$ units, then $AB=2x$ units
Using Pythagoras theorem,
$⇒AC^2=AB^2+BC^2$
$⇒AC^2=x^2+(2x)^2$
$⇒AC^2=5x^2$
$⇒AC =x\sqrt5 $ units
Now, we have $\triangle ABC$ right-angled at B, with $\tan A=\frac{1}{2}$, and we've found that $BC=x$, $AB=2x$, and $AC=x\sqrt{5}$.
$⇒\sin A = \frac{BC}{AC} = \frac{x}{x\sqrt{5}} = \frac{1}{\sqrt{5}}$
$⇒\cos A = \frac{AB}{AC} = \frac{2x}{x\sqrt{5}} = \frac{2}{\sqrt{5}}$
$⇒\sin C = \frac{AB}{AC} = \frac{2x}{x\sqrt{5}} = \frac{2}{\sqrt{5}}$
$⇒\cos C = \frac{BC}{AC} = \frac{x}{x\sqrt{5}} = \frac{1}{\sqrt{5}}$
Substituting these values into the given expression, we get:
$⇒\frac{\sin A(\cos C+\cos A)}{\cos C(\sin C-\sin A)} = \frac{\frac{1}{\sqrt{5}}(\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}})}{\frac{1}{\sqrt{5}}(\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{5}})} = \frac{\frac{3}{5}}{\frac{1}{5}} = 3$
Hence, the correct answer is 3.
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