10th Class Mathematics Introduction to Trigonometry Question Bank MCQs - Introduction to Trigonometry

  • question_answer
    If \[\tan \theta =\frac{1}{\sqrt{7}},\]then \[\frac{(\text{cose}{{\text{c}}^{2}}\theta -{{\sec }^{2}}\theta )}{(\text{cose}{{\text{c}}^{2}}\theta +{{\sec }^{2}}\theta )}=\]

    A) \[\frac{-2}{3}\]

    B) \[\frac{-3}{4}\]

    C) \[\frac{2}{3}\]

    D) \[\frac{3}{4}\]

    Correct Answer: D

    Solution :

    [d] We have, \[\tan \theta =\frac{1}{\sqrt{7}}\,\,\,\,\Rightarrow \,\,\,\,\cot \theta =\sqrt{7}\]
    We know,
    \[{{\sec }^{2}}\theta =(1+{{\tan }^{2}}\theta )=\left( 1+{{\left( \frac{1}{\sqrt{7}} \right)}^{2}} \right)=\left( 1+\frac{1}{7} \right)=\frac{8}{7}\]
    and  \[\cos e{{c}^{2}}\theta =(1+{{\cot }^{2}}\theta )=(1+{{(\sqrt{7})}^{2}})=(1+7)=8\] 
    \[\therefore \,\,\,\,\,\frac{\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta }{\cos e{{c}^{2}}\theta +{{\sec }^{2}}\theta }=\frac{\left( 8-\frac{8}{7} \right)}{\left( 8+\frac{8}{7} \right)}=\frac{48}{64}=\frac{3}{4}\]


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