10th Class Mathematics Introduction to Trigonometry Question Bank Trigonometry

  • question_answer
    In the adjoining figure the value of \[\mathbf{secl}{{\mathbf{7}}^{{}^\circ }}-\mathbf{sm7}{{\mathbf{3}}^{{}^\circ }}\]is

    A) \[\frac{y}{x\sqrt{{{y}^{2}}-{{x}^{2}}}}\]

    B) \[\frac{{{x}^{2}}}{y\sqrt{{{y}^{2}}-{{x}^{2}}}}\]

    C) \[\frac{{{x}^{2}}}{y\sqrt{{{x}^{2}}-{{y}^{2}}}}\]

    D) \[\frac{{{y}^{2}}}{x\sqrt{{{x}^{2}}-{{y}^{2}}}}\]

    Correct Answer: B

    Solution :

    (b): \[\text{sin}{{17}^{{}^\circ }}=\frac{x}{y}\] \[cos{{17}^{{}^\circ }}=\sqrt{1-si{{n}^{2}}{{17}^{{}^\circ }}}\] \[=\sqrt{1-\frac{{{x}^{2}}}{{{y}^{2}}}}=\sqrt{\frac{{{y}^{2}}-{{x}^{2}}}{{{y}^{2}}}}\] \[\sqrt{\frac{{{y}^{2}}-{{x}^{2}}}{{{y}^{2}}}}\] \[\therefore sec{{17}^{{}^\circ }}=\frac{y}{\sqrt{{{y}^{2}}-{{x}^{2}}}}\] \[sin{{73}^{{}^\circ }}=sin\left( {{90}^{{}^\circ }}-{{17}^{{}^\circ }} \right)=cos{{17}^{{}^\circ }}\] \[\therefore sec{{17}^{{}^\circ }}-sin{{73}^{{}^\circ }}\] \[=\frac{y}{\sqrt{{{y}^{2}}-{{x}^{2}}}}-\frac{\sqrt{{{y}^{2}}-{{x}^{2}}}}{y}\] \[=\frac{{{y}^{2}}-{{y}^{2}}+{{x}^{2}}}{y\sqrt{{{y}^{2}}-{{x}^{2}}}}=\frac{{{x}^{2}}}{y\sqrt{{{y}^{2}}-{{x}^{2}}}}\]


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