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JEE Main & Advanced Mathematics Trigonometric Equations Question Bank Relation between sides and angles, Solutions of triangles

  • question_answer
    If in triangle \[ABC,\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}=\frac{\sin (A-B)}{\sin (A+B)}\], then the triangle is  [Roorkee 1987]

    A) Right angled

    B) Isosceles

    C) Right angled or isosecles

    D) Right angled isosecles

    Correct Answer: C

    Solution :

    \[\frac{\sin (A-B)}{\sin (A+B)}=\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\] Þ  \[\frac{\sin (A-B)}{\sin (A+B)}=\frac{{{\sin }^{2}}A-{{\sin }^{2}}B}{{{\sin }^{2}}A+{{\sin }^{2}}B}\] Þ  \[\frac{\sin (A-B)}{\sin C}=\frac{\sin (A-B)\sin (A+B)}{{{\sin }^{2}}A+{{\sin }^{2}}B}\] Þ  \[\sin (A-B)\text{ }\left[ \frac{1}{\sin C}-\frac{\sin C}{{{\sin }^{2}}A+{{\sin }^{2}}B} \right]=0\] Either \[\sin (A-B)=0\Rightarrow A=B\] i.e. isosceles or \[{{\sin }^{2}}A+{{\sin }^{2}}B={{\sin }^{2}}C\] or \[{{a}^{2}}+{{b}^{2}}={{c}^{2}}\] i.e., right angled triangle.


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