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Q.
If $\sec \, \theta - \tan \, \theta = \frac{1}{2}, $ then $\theta$ lies in
Trigonometric Functions
Solution:
Given $\sec \, \theta - \tan \, \theta = \frac{1}{2}$ ....(1)
Also $\sec^2 \, \theta - \tan^2 \, \theta = 1 $ ....(2)
Dividing (2) by (1), we get
$\sec \, \theta + \tan \, \theta = 2$ ....(3)
Adding (1) and (3), we get
$2 \, \sec \, \theta = \frac{5}{2} \, \Rightarrow \, \sec \, \theta = \frac{5}{4}$
and subtracting (2) from (1), we get
$2 \tan \, \theta = \frac{3}{2} \, \Rightarrow \, \tan \, \theta = \frac{3}{4}$
Since both sec $\theta$ and tan $\theta$ are positive, therefore, $\theta$ lies in the first quadrant.