2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-31

  • question_answer
    In triangle ABC, if \[\Delta ={{a}^{2}}-{{({{b}^{2}}-c)}^{2}}\] then tan A is equal to [Note: All symbols used have usual meaning in triangle ABC.]

    A) \[\frac{7}{15}\]                                

    B) \[\frac{8}{15}\]

    C) \[\frac{11}{15}\]                                              

    D) \[\frac{13}{15}\]

    Correct Answer: B

    Solution :

    Given, \[\Delta =(a+b+c)(a+c-b)\]                 \[\Rightarrow \Delta =2(s-c).2(s-b)\]                      ?(1) Also, \[\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\frac{(s-b)(s-c)}{\Delta }=\frac{1}{4}\] So, \[\tan A=\frac{2\tan \frac{A}{2}}{1-{{\tan }^{2}}\frac{A}{2}}=\frac{2\times \frac{1}{4}}{1-{{\left( \frac{1}{4} \right)}^{2}}}=\frac{8}{15}\]


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