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

  • question_answer
    Match the following.
    Column - I Column- II
    P. \[\frac{\sin \,0{}^\circ }{\cos 90{}^\circ }+\sin \,45{}^\circ \] 1. \[\left( \frac{1-\sqrt{3}}{2} \right)\]
    Q. \[\cos \,60{}^\circ -\sin \,60{}^\circ \] 2. \[1+\frac{\sqrt{2}}{3}\]
    R. \[\sec 30{}^\circ \sin 60{}^\circ \] \[+\cos 45{}^\circ \cos ec\,60{}^\circ \] 3. 1
    S. \[\frac{{{\cos }^{3}}30{}^\circ -{{\cos }^{3}}60{}^\circ }{{{\sin }^{3}}60{}^\circ -{{\sin }^{3}}30{}^\circ }\] 4. \[\frac{1}{\sqrt{2}}\]

    A) P-2,      Q-4,      R-3,      S-1

    B) P-3,      Q-1,      R-4,      S-1

    C) P-2,      Q-3,      R-4,      S-1

    D) P-4,      Q-1,      R-2,      S-3

    Correct Answer: D

    Solution :

    \[P\to 4;\,Q\to 1;\,R\to 2;\,S\to 3\] (P) \[\frac{\sin \,0{}^\circ }{\cos 90{}^\circ }+\sin 45{}^\circ =0+\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\] (Q) \[\cos 60{}^\circ -\sin 60{}^\circ =\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2}\] (R) \[\sec \,30{}^\circ \sin 60{}^\circ +\cos 45{}^\circ \cos ec60{}^\circ \] \[=\frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{2}{\sqrt{3}}=1+\frac{\sqrt{2}}{\sqrt{3}}\] (S) \[\frac{{{\cos }^{3}}30{}^\circ -{{\cos }^{3}}60{}^\circ }{{{\sin }^{3}}60{}^\circ -{{\sin }^{3}}30{}^\circ }\] \[=\frac{{{\left( \frac{\sqrt{3}}{2} \right)}^{3}}-{{\left( \frac{1}{2} \right)}^{3}}}{{{\left( \frac{\sqrt{3}}{2} \right)}^{3}}-{{\left( \frac{1}{2} \right)}^{3}}}=1\]


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