JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Trigonometrical ratios of multiple and sub multiple angles

  • question_answer
    If \[\sin \theta +\sin \varphi =a\]and \[\cos \theta +\cos \varphi =b,\]then \[\tan \frac{\theta -\varphi }{2}\]is equal to [MP PET 1993]

    A) \[\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{4-{{a}^{2}}-{{b}^{2}}}}\]

    B) \[\sqrt{\frac{4-{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}\]

    C) \[\sqrt{\frac{{{a}^{2}}+{{b}^{2}}}{4+{{a}^{2}}+{{b}^{2}}}}\]

    D) \[\sqrt{\frac{4+{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}\]

    Correct Answer: B

    Solution :

    Given that \[\sin \theta +\sin \varphi =a\] ?..(i) and \[\cos \theta +\cos \varphi =b\] ?..(ii) Squaring, \[{{\sin }^{2}}\theta +{{\sin }^{2}}\varphi +2\sin \theta \sin \varphi ={{a}^{2}}\] and \[{{\cos }^{2}}\theta +{{\cos }^{2}}\varphi +2\cos \theta \cos \varphi ={{b}^{2}}\] Adding, 2+ 2 \[(\sin \theta \sin \varphi +\cos \theta \cos \varphi )={{a}^{2}}+{{b}^{2}}\] Þ\[2\cos (\theta -\varphi )={{a}^{2}}+{{b}^{2}}-2\]Þ  \[\cos (\theta -\varphi )=\frac{{{a}^{2}}+{{b}^{2}}-2}{2}\] \[\Rightarrow \frac{1-{{\tan }^{2}}\frac{\theta -\varphi }{2}}{1+{{\tan }^{2}}\frac{\theta -\varphi }{2}}=\frac{{{a}^{2}}+{{b}^{2}}-2}{2}\] Þ \[({{a}^{2}}+{{b}^{2}})+({{a}^{2}}+{{b}^{2}}){{\tan }^{2}}\frac{\theta -\varphi }{2}-2-2{{\tan }^{2}}\frac{\theta -\varphi }{2}\] \[=2-2{{\tan }^{2}}\frac{\theta -\varphi }{2}\] Þ\[\frac{4-{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}={{\tan }^{2}}\frac{\theta -\varphi }{2}\]Þ \[\tan \frac{(\theta -\varphi )}{2}=\sqrt{\frac{4-{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}}\]  Trick: Put\[\theta =\frac{\pi }{2},\varphi ={{0}^{o}}\], then \[a=1=b\] \\[\tan \frac{\theta -\varphi }{2}=1\], which is given by (a) and (b). Again putting\[\theta =\frac{\pi }{4}=\varphi \], we get\[\tan \frac{\theta -\varphi }{2}=0\], which is given by (b).

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