2. Questions Bank
  3. 10th Class
  4. Mathematics
  5. Introduction to Trigonometry

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

  • question_answer
    \[(1+{{\tan }^{2}}A)+\left( 1+\frac{1}{{{\tan }^{2}}A} \right)\] is equal to

    A) \[\frac{1}{{{\sin }^{2}}A-{{\sin }^{4}}A}\]       

    B) \[\frac{1}{{{\sin }^{2}}A+{{\sin }^{4}}A}\]

    C)           \[\frac{{{\cos }^{2}}A}{\sin A+{{\sin }^{2}}A}\]

    D)        \[\frac{{{\cos }^{2}}A}{\sin A-{{\sin }^{2}}A}\]

    Correct Answer: A

    Solution :

    We have: \[(1+{{\tan }^{2}}A)+\left( 1+\frac{1}{{{\tan }^{2}}A} \right)\] \[={{\sec }^{2}}A+(1+{{\cot }^{2}}A)\] \[[\therefore \,\,1+ta{{n}^{2}}A={{\sec }^{2}}A]\] \[={{\sec }^{2}}A+\cos e{{c}^{2}}A\] \[[\therefore \,\,1+co{{t}^{2}}A=co{{\sec }^{2}}A]\] \[=\frac{1}{co{{s}^{2}}A}+\frac{1}{{{\sin }^{2}}A}=\frac{1}{{{\sin }^{2}}A(1-{{\sin }^{2}}A)}\] \[=\frac{1}{{{\sin }^{2}}A-{{\sin }^{4}}A}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner