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10th Class Mathematics Introduction to Trigonometry Question Bank Trigonometry

  • question_answer
    If \[\frac{1+\sin \alpha }{1-\sin \alpha }=\frac{{{m}^{2}}}{{{n}^{2}}},\] then sin \[\alpha \] is:

    A)  \[\frac{{{m}^{2}}+{{n}^{2}}}{{{m}^{2}}-{{n}^{2}}}\]              

    B)  \[\frac{{{m}^{2}}-{{n}^{2}}}{{{m}^{2}}+{{n}^{2}}}\]

    C)  \[\frac{{{m}^{2}}+{{n}^{2}}}{{{n}^{2}}+{{m}^{2}}}\]            

    D)  \[\frac{{{n}^{2}}-{{m}^{2}}}{{{m}^{2}}+{{n}^{2}}}\]

    Correct Answer: B

    Solution :

    (b): \[{{n}^{2}}(1+sin\alpha )={{m}^{2}}(1-sin\alpha )\] \[\Rightarrow \]   \[{{n}^{2}}+{{n}^{2}}\sin \alpha ={{m}^{2}}-{{m}^{2}}\sin \alpha \] \[\Rightarrow \]   \[\left( {{m}^{2}}+{{n}^{2}} \right)\sin \alpha ={{m}^{2}}-{{n}^{{}}}\] \[\Rightarrow \]   \[\sin \alpha =\frac{{{m}^{2}}-{{n}^{2}}}{{{m}^{2}}+{{n}^{2}}}\]


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