| If \[\tan \theta \cdot \tan 2\theta =1,\]then the value of \[{{\sin }^{2}}2\theta +{{\tan }^{2}}2\theta \] is equal to [SSC (CGL) 2011] |
A) \[\frac{3}{4}\]
B) \[\frac{10}{3}\]
C) \[3\frac{3}{4}\]
D) \[8\]
Correct Answer: C
Solution :
| \[\tan \theta \cdot \tan 2\theta =1\] |
| \[\Rightarrow \]\[\tan 2\theta =\frac{1}{\tan \theta }=\cot \theta =\tan (90-\theta )\] |
| \[\Rightarrow \]\[2\theta =90-\theta \]\[\Rightarrow \]\[3\theta =90{}^\circ \] |
| \[\theta =30{}^\circ \] |
| \[\therefore \]\[{{\sin }^{2}}2\theta +{{\tan }^{2}}2\theta ={{\sin }^{2}}(60{}^\circ )+{{\tan }^{2}}(60{}^\circ )\] |
| \[={{\left( \frac{\sqrt{3}}{2} \right)}^{2}}+{{(\sqrt{3})}^{2}}\] |
| \[=\frac{3}{4}+3=\frac{15}{4}=3\frac{3}{4}\] |
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