question_answer

If \[\tan \theta =\frac{1}{\sqrt{7}},\]then\[\frac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta +{{\sec }^{2}}\theta }\]is equal to

A)  \[\frac{-3}{4}\]

B)         \[\frac{-2}{3}\]

C)  \[\frac{2}{3}\]             

D)  \[\frac{3}{4}\]

Correct Answer: D

Solution :

Given, \[\tan \theta =\frac{1}{\sqrt{7}}\] \[AC=\sqrt{A{{B}^{2}}+C{{B}^{2}}}=\sqrt{1+7}=2\sqrt{2}\] Then, \[\frac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta +{{\sec }^{2}}\theta }\]?(i) \[\because \]From triangle ABC \[\text{cosec}\theta =\frac{2\sqrt{2}}{1}\]and \[\sec \theta =\frac{2\sqrt{2}}{\sqrt{7}}\] Put the value of s\[\sec \theta \]and\[\cos ec\theta \]in Eq. (i) \[\Rightarrow \]\[\frac{{{(2\sqrt{2})}^{2}}-{{\left( \frac{2\sqrt{2}}{\sqrt{7}} \right)}^{2}}}{{{(2\sqrt{2})}^{2}}+{{\left( \frac{2\sqrt{2}}{\sqrt{7}} \right)}^{2}}}=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}\] \[=\frac{56-8}{56+8}=\frac{48}{64}=\frac{3}{4}\]