question_answer

If \[\theta =\frac{1}{2},\]then \[\frac{\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta }{\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta }\]is equal to

A) \[\frac{3}{5}\]  

B) \[\frac{1}{5}\]

C) \[\frac{2}{5}\]              

D) \[\frac{4}{5}\]

Correct Answer: A

Solution :

\[\tan \theta =\frac{1}{2}=\frac{BC}{AB}\]

\[\therefore \]\[BC=1\] and \[AB=2\]

Now, \[{{(AC)}^{2}}={{(AB)}^{2}}+{{(BC)}^{2}}\]

\[\Rightarrow \]   \[{{(AC)}^{2}}={{(2)}^{2}}+{{(1)}^{2}}\]

\[\Rightarrow \]   \[AC=\sqrt{5}\]

\[\therefore \]\[\operatorname{cosec}\theta =\frac{AC}{BC}=\sqrt{5}\]

and \[\sec \theta =\frac{AC}{AB}=\frac{\sqrt{5}}{2}\]

\[\therefore \]\[\frac{\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta }{\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta }=\frac{{{(\sqrt{5})}^{2}}-{{\left( \frac{\sqrt{5}}{2} \right)}^{2}}}{{{(\sqrt{5})}^{2}}+{{\left( \frac{\sqrt{5}}{2} \right)}^{2}}}\]

\[=\frac{5-\frac{5}{4}}{5+\frac{5}{4}}=\frac{15}{25}=\frac{3}{5}\]