JEE Main & Advanced Mathematics Trigonometric Identities Question Bank Trigonometrical ratios of sum and difference of two and three angles

  • question_answer
    If \[m\tan (\theta -30{}^\circ )=n\tan (\theta +120{}^\circ ),\] then \[\frac{m+n}{m-n}=\] [IIT 1966]

    A) \[2\,\cos \,2\theta \]

    B) \[\cos \,\,2\theta \]

    C) \[2\,\sin \,2\theta \]

    D) \[\sin \,\,2\theta \]

    Correct Answer: A

    Solution :

    \[\frac{m}{n}=\frac{\tan \,({{120}^{o}}+\theta )}{\tan \,(\theta -{{30}^{o}})}\] \[\Rightarrow \,\,\frac{m+n}{m-n}=\frac{\tan \,(\theta +{{120}^{o}})+\tan \,(\theta -{{30}^{o}})}{\tan \,(\theta +{{120}^{o}})-\tan \,(\theta -{{30}^{o}})}\] (By componendo and dividendo) \[=\frac{\sin (\theta +{{120}^{o}})\cos (\theta -{{30}^{o}})+\cos (\theta +{{120}^{o}})\sin (\theta -{{30}^{o}})}{\sin (\theta +{{120}^{o}})\cos (\theta -{{30}^{o}})-\cos (\theta +{{120}^{o}})\sin (\theta -{{30}^{o}})}\] \[=\frac{\sin \,(2\theta +{{90}^{o}})}{\sin \,({{150}^{o}})}=\frac{\cos \,2\theta }{1/2}=2\,\cos \,2\theta \].


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