A) \[\frac{-3}{4}\]
B) \[\frac{-2}{3}\]
C) \[\frac{2}{3}\]
D) \[\frac{3}{4}\]
Correct Answer: D
Solution :
Given, \[\tan \theta =1\] \[AC=\sqrt{A{{B}^{2}}+C{{B}^{2}}}\] \[=\sqrt{1+7}=2\sqrt{2}\] Then, \[\frac{\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta }{\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta }\] ?(i) From \[\Delta ABC,\] \[\text{cosec}\theta =\frac{2\sqrt{2}}{1}\] and \[\sec \theta =\frac{2\sqrt{2}}{\sqrt{7}}\] Put the value of \[\sec \theta \] and\[\text{cosec}\theta \] in Eq. (i) \[\Rightarrow \]\[\frac{{{(2\sqrt{2})}^{2}}-{{\left( \frac{2\sqrt{2}}{\sqrt{7}} \right)}^{2}}}{{{(2\sqrt{2})}^{2}}+{{\left( \frac{2\sqrt{2}}{7} \right)}^{2}}}=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}\] \[=\frac{56-8}{56+8}=\frac{48}{64}=\frac{3}{4}\]
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