Question

If $\sec \theta +\tan \theta =\frac{2}{3}$, find the value of $\sin \theta$ and determine the quadrant in which $\theta$ lies.

Answer 1

$sec\theta +tan\theta =\frac{2}{3}$ ............... (1)

$se{c}^2\theta -ta{n}^2\theta =1$

$(sec\theta +tan\theta )(sec\theta -tan\theta )=1$

$\Rightarrow sec\theta -tan\theta =\frac{1}{sec\theta +tan\theta }$

$sec\theta -tan\theta =\frac{3}{2}$ ................... (2)

Add equation (1) and (2),

$2sec\theta =\frac{2}{3}+\frac{3}{2}=\frac{13}{6}$

$sec\theta =\frac{13}{12}$

From (1),

$tan\theta =\frac{2}{3}-sec\theta =-\frac{5}{12}$

$sin\theta =\frac{tan\theta }{sec\theta }=-\frac{5}{13}$

$sin\theta$ and $tan\theta$ are negative, so $\theta$ lies in fourth quadrant.