Question : If $\tan(\alpha -\beta)=1,\sec(\alpha +\beta)=\frac{2}{\sqrt{3}}$ and $\alpha ,\beta$ are positive, then the smallest value of $\alpha$ is:
Option 1: $142\frac{1}{2}°$
Option 2: $187\frac{1}{2}°$
Option 3: $7\frac{1}{2}°$
Option 4: $37\frac{1}{2}°$
Correct Answer: $37\frac{1}{2}°$
Solution :
Given:
$\tan(\alpha -\beta)=1$
$\sec(\alpha +\beta)=\frac{2}{\sqrt{3}}$
Formula Used:
$\tan 45° = 1$
$\sec 45° = \frac{2}{\sqrt{3}}$
Calculation:
$\tan(\alpha -\beta)=1$
⇒ $\alpha -\beta = 45°$..............(i)
Also, $\sec(\alpha +\beta)=\frac{2}{\sqrt{3}}$
⇒ $\alpha +\beta = 30°$................(ii)
Adding (i) + (ii)
$2\alpha = 45° + 30°$
⇒ $\alpha= \frac{75}{2} = 37\frac{1}{2}°$
Hence, the correct answer is $37\frac{1}{2}°$.
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