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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