8th Class Mathematics Introduction to Trigonometry Question Bank Trigonometry

  • question_answer If \[se{{c}^{2}}\theta +ta{{n}^{2}}\theta =\sqrt{2}\], then the value of \[\left( \mathbf{se}{{\mathbf{c}}^{\mathbf{4}}}\theta -\mathbf{ta}{{\mathbf{n}}^{\mathbf{4}}}\theta  \right)\] is

    A)  \[\frac{1}{\sqrt{3}}\]                          

    B)  1

    C)  \[\sqrt{2}\] 

    D)  0

    Correct Answer: C

    Solution :

    (c): \[se{{c}^{2}}\theta +ta{{n}^{2}}\theta =\sqrt{2}\] And \[se{{c}^{2}}\theta -{{\tan }^{2}}\theta =1\] \[\therefore se{{c}^{4}}\theta -ta{{n}^{4}}\theta \] = (sec2 9 + tan2 9)(sec2 9 - tan2 9) \[=\sqrt{2}+1=\sqrt{2}\]        

adversite


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

Free
Videos