A) - 4/5 but not 4/5
B) - 4/5 or 4/5
C) 4/5 but not - 4/5
D) None of these
Correct Answer: B
Solution :
Since \[\text{cose}{{\text{c}}^{2}}\theta =1+{{\cot }^{2}}\theta =1+\frac{9}{16}=\frac{25}{16}\] \[\left( \because \tan \theta =-\frac{4}{3} \right)\] \[{{\sin }^{2}}\theta =\frac{1}{\text{cose}{{\text{c}}^{2}}\theta }=\frac{16}{25}\Rightarrow \sin \theta =\pm \frac{4}{5},\] Both the values are acceptable, since \[\tan \theta =-\frac{4}{3}\,\,\] \[\,i.e.,\theta \] lies in \[{{2}^{nd}}\]or \[{{4}^{th}}\]quadrant.
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