10th Class Mathematics Introduction to Trigonometry Question Bank Trigonometry

  • question_answer
    \[\frac{\tan \theta }{{{(1+{{\tan }^{2}}\theta )}^{2}}}+\frac{cot\theta }{{{(1+{{\cot }^{2}}\theta )}^{2}}}\] is equal to

    A) \[2\sin \theta \,\,.\,\,\cos \theta \]

    B) \[\operatorname{cosec}\theta \,\,.\,\,sec\theta \]

    C) \[\sin \,\theta \,\,.\,\,cos\theta \]

    D) \[2cosec\,\theta \,\,.\,\,sin\theta \]

    Correct Answer: C

    Solution :

     \[\frac{\tan \theta }{{{(1+{{\tan }^{2}}\theta )}^{2}}}+\frac{\cot \theta }{{{(1+{{\cot }^{2}}\theta )}^{2}}}=\frac{\tan \theta }{se{{c}^{4}}\theta }+\frac{\cot \theta }{\text{cose}{{\text{c}}^{4}}\theta }\] \[[\because \,{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta ,\text{cose}{{\text{c}}^{2}}\theta =1+{{\cot }^{2}}\theta ]\]                 \[=\sin \theta \,{{\cos }^{3}}\theta +\cos \theta {{\sin }^{3}}\theta \]                 \[=\sin \theta \cos \theta ({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )\]                 \[=\sin \theta \,\cos \theta \]


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