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 \[\sec \theta =\frac{AC}{AB}=\frac{\sqrt{5}}{2}\] |
\[\therefore \]\[\frac{\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta }{\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta }=\frac{{{(\sqrt{5})}^{2}}-{{\left( \frac{\sqrt{5}}{2} \right)}^{2}}}{{{(\sqrt{5})}^{2}}+{{\left( \frac{\sqrt{5}}{2} \right)}^{2}}}\] |
\[=\frac{5-\frac{5}{4}}{5+\frac{5}{4}}=\frac{15}{25}=\frac{3}{5}\] |
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