Question : In $\triangle ABC, \angle B=90^{\circ}$ and AB : BC = 1 : 2. The value of $\cos A+\tan C$ is:
Option 1: $\frac{5+\sqrt{5}}{2 \sqrt{5}}$
Option 2: $\frac{1+\sqrt{5}}{2 \sqrt{5}}$
Option 3: $\frac{2 \sqrt{5}}{2+\sqrt{5}}$
Option 4: $\frac{2+\sqrt{5}}{2 \sqrt{5}}$
Correct Answer: $\frac{2+\sqrt{5}}{2 \sqrt{5}}$
Solution :
In $\triangle ABC, \angle B=90^{\circ}$
AB : BC = 1 : 2
⇒ $\frac{\text{AB}}{\text{BC}} = \frac{1}{2}$
⇒ $\frac{\text{AB}}{1} = \frac{\text{BC}}{2}=$ k
⇒ AB = k, BC = 2k
Using the Pythagoras theorem,
AC$^2$ = BC$^2$ + AB$^2$ = (2k)$^2$ + k$^2$ = 5k$^2$
So, AC = $\sqrt{5}$k
By definition of trigonometric ratios,
$\cos A = \frac{\text{AB}}{\text{AC}} = \frac{1}{\sqrt{5}}$
$\tan C = \frac{\text{AB}}{\text{BC}} = \frac{1}{2}$
⇒ $\cos A+\tan C$ = $ \frac{1}{2}+ \frac{1}{\sqrt{5}}$ = $\frac{\sqrt{5}+2}{2 \sqrt{5}}$
Hence, the correct answer is $\frac{2+\sqrt{5}}{2 \sqrt{5}}$.
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