10th Class Mathematics Introduction to Trigonometry Question Bank Trigonometry

  • question_answer
    In the given figure the value of \[\frac{1+2\sin \theta .\cos \theta }{1-2\sin \theta .\cos \theta }\] is

    A) 810

    B) 540

    C) 900

    D) 490

    Correct Answer: C

    Solution :

    (c): \[29tan\theta =31\Rightarrow tan\theta =\frac{31}{29}\] Expression= \[\frac{1+2sin\theta .cos\theta }{1-2\sin cos\theta }\] \[=\frac{si{{n}^{2}}\theta +co{{s}^{2}}\theta +2\sin \theta .cos\theta }{si{{n}^{2}}\theta +co{{s}^{2}}\theta -2sin.cos\theta }\] \[=\frac{{{\left( sin\theta +cos\theta  \right)}^{2}}}{{{\left( sin\theta -cos\theta  \right)}^{2}}}\] \[=\left( \frac{\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\cos \theta }}{\frac{\sin \theta }{\cos \theta }-\frac{\cos \theta }{\cos \theta }} \right)={{\left( \frac{\tan \theta +1}{\tan \theta -1} \right)}^{2}}\] \[={{\left( \frac{\frac{31}{29}+1}{\frac{31}{29}-1} \right)}^{2}}={{\left( \frac{\frac{31+29}{29}}{\frac{31-29}{29}} \right)}^{2}}\] \[={{\left( \frac{60}{2} \right)}^{2}}{{\left( 30 \right)}^{2}}=900\]


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