Question : If $\sec \beta+\tan \beta=2$, then what is the value of $\cot \beta$?
Option 1: $\frac{5}{3}$
Option 2: $\frac{3}{5}$
Option 3: $\frac{4}{3}$
Option 4: $\frac{3}{4}$
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Correct Answer: $\frac{4}{3}$
Solution : Given: $\sec \beta+\tan \beta=2$..........(1) We know that $\sec^2 \beta-\tan^2 \beta=1$ ⇒ $(\sec \beta-\tan \beta)(\sec \beta+\tan \beta)=1$ ⇒ $(\sec \beta-\tan \beta)=\frac{1}{2}$...............(2) Subtracting equation (2) from equation (1), we get: ⇒ $2\tan\beta=2-\frac{1}{2}=\frac{3}{2}$ ⇒ $\tan\beta= \frac{3}{4}$ $\therefore \cot\beta=\frac{4}{3}$ Hence, the correct answer is $\frac{4}{3}$.
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