Question:
If \(m\cos(\alpha + \beta) - n\cos(\alpha - \beta) = m\cos(\alpha - \beta) + n\cos(\alpha + \beta)\), then \(\tan \alpha \tan \beta\) is:
If \(m\cos(\alpha + \beta) - n\cos(\alpha - \beta) = m\cos(\alpha - \beta) + n\cos(\alpha + \beta)\), then \(\tan \alpha \tan \beta\) is:
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For equations involving trigonometric identities, always consider rewriting them using basic sine and cosine formulas to facilitate the process.
Updated On: Mar 6, 2026
- \(m + n\)
- \(m - n\)
- \(-\frac{n}{m}\)
- \(\frac{m}{n}\)
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The Correct Option is C
Solution and Explanation
Step 1: Simplify the given equation.
\[ m\cos(\alpha + \beta) - n\cos(\alpha - \beta) = m\cos(\alpha - \beta) + n\cos(\alpha + \beta) \] \[ m\cos(\alpha + \beta) - m\cos(\alpha - \beta) = n\cos(\alpha + \beta) + n\cos(\alpha - \beta) \] \[ m[\cos(\alpha + \beta) - \cos(\alpha - \beta)] = n[\cos(\alpha + \beta) + \cos(\alpha - \beta)] \] Step 2: Apply trigonometric identities for sum and difference of angles.
Using \(\cos(\alpha + \beta) - \cos(\alpha - \beta) = -2\sin \alpha \sin \beta\) and \(\cos(\alpha + \beta) + \cos(\alpha - \beta) = 2\cos \alpha \cos \beta\): \[ -2m\sin \alpha \sin \beta = 2n\cos \alpha \cos \beta. \] \[ \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = -\frac{n}{m}. \] Step 3: Conclude with the value of \(\tan \alpha \tan \beta\).
\[ \tan \alpha \tan \beta = \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = -\frac{n}{m}. \]
\[ m\cos(\alpha + \beta) - n\cos(\alpha - \beta) = m\cos(\alpha - \beta) + n\cos(\alpha + \beta) \] \[ m\cos(\alpha + \beta) - m\cos(\alpha - \beta) = n\cos(\alpha + \beta) + n\cos(\alpha - \beta) \] \[ m[\cos(\alpha + \beta) - \cos(\alpha - \beta)] = n[\cos(\alpha + \beta) + \cos(\alpha - \beta)] \] Step 2: Apply trigonometric identities for sum and difference of angles.
Using \(\cos(\alpha + \beta) - \cos(\alpha - \beta) = -2\sin \alpha \sin \beta\) and \(\cos(\alpha + \beta) + \cos(\alpha - \beta) = 2\cos \alpha \cos \beta\): \[ -2m\sin \alpha \sin \beta = 2n\cos \alpha \cos \beta. \] \[ \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = -\frac{n}{m}. \] Step 3: Conclude with the value of \(\tan \alpha \tan \beta\).
\[ \tan \alpha \tan \beta = \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = -\frac{n}{m}. \]
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