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If 3tanθ = 2√3sin, 0° < θ < 90°, then the value of $$\frac{{{\text{cose}}{{\text{c}}^2}2\theta + {{\cot }^2}2\theta }}{{{{\sin }^2}\theta + {{\tan }^2}\theta }}$$ is:
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& 3\tan \theta = 2\sqrt 3 \sin \theta \cr
& \cos \theta = \frac{{\sqrt 3 }}{2} = \frac{B}{H} \cr
& P = \sqrt {{3^2} + {2^2}} = \sqrt {13} \cr
& \cos \theta = \cos {30^ \circ } \cr
& \theta = {30^ \circ } \cr
& \Rightarrow \frac{{{\text{cose}}{{\text{c}}^2}2\theta + {{\cot }^2}2\theta }}{{{{\sin }^2}\theta + {{\tan }^2}2\theta }} \cr
& = \frac{{{\text{cose}}{{\text{c}}^2}{{60}^ \circ } + {{\cot }^2}{{60}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {{\tan }^2}{{60}^ \circ }}} \cr
& = \frac{{\frac{4}{3} + \frac{1}{3}}}{{\frac{1}{4} + 3}} \cr
& = \frac{{\frac{5}{3}}}{{\frac{{13}}{4}}} \cr
& = \frac{{20}}{{39}} \cr} $$
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