10th Class Mathematics Introduction to Trigonometry Question Bank Case Based (MCQs) - Introduction to Trigonometry

  • question_answer
    Value of \[{{\sin }^{2}}S-{{\cos }^{2}}R\] is:

    A) \[0\]

    B) \[1\]

    C) \[\frac{97}{85}\]

    D) \[\frac{589}{986}\]

    Correct Answer: D

    Solution :

    From part (1), \[\sec S=\frac{\sqrt{34}}{3}\Rightarrow \cos S=\frac{3}{\sqrt{34}}\]
    \[\therefore \,\,\,\,\sin S=\sqrt{1-{{\cos }^{2}}S}=\sqrt{1-\frac{9}{34}}=\sqrt{\frac{25}{34}}=\frac{5}{\sqrt{34}}\]
    From part (2), \[\cos ec\,R=\frac{\sqrt{29}}{5}\Rightarrow \sin R=\frac{5}{\sqrt{29}}\]
    \[\therefore \,\,\,\,\,\,\,\,\,\cos R=\sqrt{1-{{\sin }^{2}}R}=\sqrt{1-\frac{25}{29}}=\sqrt{\frac{4}{29}}=\frac{2}{\sqrt{29}}\]
    \[\therefore \,\,\,\,\,\,{{\sin }^{2}}S-{{\cos }^{2}}R={{\left( \frac{5}{\sqrt{34}} \right)}^{2}}-{{\left( \frac{2}{\sqrt{29}} \right)}^{2}}=\frac{25}{34}-\frac{4}{29}\]
    \[=\frac{725-136}{986}=\frac{589}{986}\]
    So, option [d] is correct.


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