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: $37\frac{1}{2}°$
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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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