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10th Class Mathematics Introduction to Trigonometry Question Bank Trigonometry

  • question_answer
    If\[\mathbf{sin}\theta -\mathbf{cos}\theta =\sqrt{2}\mathbf{sin}\left( \mathbf{9}{{\mathbf{0}}^{{}^\circ }}-\theta  \right)\]then \[\mathbf{cot}\theta \]is equal to:

    A) \[\sqrt{2}\]

    B) 0

    C) \[\sqrt{2}-1\]

    D) \[\sqrt{2}+1\]

    Correct Answer: C

    Solution :

    (c): \[sin\theta -cos\theta =\sqrt{2}sin({{90}^{{}^\circ }}-\theta )\] \[sin\theta -cos\theta =\sqrt{2}cos\theta \] \[\Rightarrow \sqrt{2}\cos \theta +cos\theta =sin\theta \] \[\Rightarrow cos\theta (\sqrt{2}+1)=sin\theta \] \[\Rightarrow \frac{cos\theta }{sin\theta }=(\sqrt{2}+1)\] \[\Rightarrow cot\theta =\frac{1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}\] \[\Rightarrow \frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1\]


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