Question : Find the value of $\tan A$, if $\sec A + \tan A = \sqrt{3}$ and $A$ lies between $0^\circ$ and $90^\circ$.
Option 1: $\sqrt{3}$
Option 2: $1$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: $0$
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Correct Answer: $\frac{1}{\sqrt{3}}$
Solution :
$\sec A + \tan A = \sqrt{3}$----------------(1)
⇒ $\frac{(\sec A + \tan A) (\sec A - \tan A)}{(\sec A - \tan A)}=\sqrt{3}$
⇒ $\frac{(\sec^2 A - \tan^2 A)}{(\sec A - \tan A)}=\sqrt{3}$ [$\because \sec^2 A - \tan^2 A = 1]$
⇒ $\frac{1}{(\sec A - \tan A)}=\sqrt{3}$
⇒ $\sec A - \tan A=\frac{1}{\sqrt{3}}$----------------(2)
Subtract equation (2) from equation (1)
$2\tan A=\sqrt3-\frac{1}{\sqrt3}$
⇒ $2\tan A=\frac{3-1}{\sqrt3}=\frac{2}{\sqrt3}$
⇒ $\tan A=\frac{1}{\sqrt3}$
Hence, the correct answer is $ \frac{1}{\sqrt{3}}$.
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