11th Class Mathematics Trigonometric Identities Question Bank Critical Thinking

  • question_answer If \[\sin A=n\sin B,\] then \[\frac{n-1}{n+1}\tan \,\frac{A+B}{2}=\]

    A) \[\sin \frac{A-B}{2}\]

    B) \[\tan \frac{A-B}{2}\]

    C) \[\cot \frac{A-B}{2}\]

    D) None of these

    Correct Answer: B

    Solution :

    We have \[\sin A=n\sin B\Rightarrow \frac{n}{1}=\frac{\sin A}{\sin B}\] \[\Rightarrow \frac{n-1}{n+1}=\frac{\sin A-\sin B}{\sin A+\sin B}=\frac{2\cos \frac{A+B}{2}\sin \frac{A-B}{2}}{2\sin \frac{A+B}{2}\cos \frac{A-B}{2}}\] \[=\tan \frac{A-B}{2}\cot \frac{A+B}{2}\] \[\Rightarrow \frac{n-1}{n+1}\tan \left( \frac{A+B}{2} \right)=\tan \frac{A-B}{2}\] .

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