If $cos( heta+\phi)=m~cos( heta-\phi)$, then $(\frac{1-m}{1+m})cot~\phi$ is equal to:

Q: If $cos(\theta+\phi)=m~cos(\theta-\phi)$, then $(\frac{1-m}{1+m})cot~\phi$ is equal to:

To solve the problem, we start with the given equation: \[ \cos(\theta + \phi) = m \cos(\theta - \phi) \] We can use the sum and difference identities for cosine: \[ \cos(\theta + \phi) = \cos\theta \cos\phi - \sin\theta \sin\phi \] \[ \cos(\theta - \phi) = \cos\theta \cos\phi + \sin\theta \sin\phi \] Substituting these into the given equation, we have: \[ \cos\theta \cos\phi - \sin\theta \sin\phi = m (\cos\theta \cos\phi + \sin\theta \sin\phi) \] Expanding the right side: \[ \cos\theta \cos\phi - \sin\theta \sin\phi = m \cos\theta \cos\phi + m \sin\theta \sin\phi \] Rearranging terms gives: \[ \cos\theta \cos\phi - m \cos\theta \cos\phi = m \sin\theta \sin\phi + \sin\theta \sin\phi \] Factor out common terms: \[ (1 - m) \cos\theta \cos\phi = (1 + m) \sin\theta \sin\phi \] Divide both sides by $\cos\theta \cos\phi$: \[ \frac{1 - m}{1 + m} = \frac{\sin\theta \sin\phi}{\cos\theta \cos\phi} \] This simplifies to: \[ \frac{1 - m}{1 + m} = \tan\theta \tan\phi \] Now, we need to find $(\frac{1-m}{1+m})\cot\phi$: \[ (\frac{1-m}{1+m})\cot\phi = \tan\theta \] Thus, the expression $(\frac{1-m}{1+m})\cot\phi$ is equal to $\tan\theta$, which corresponds to Option 1. **Explanation for Other Options:** - **Option 2: $-\tan\theta$**: This option suggests that the expression is negative, which contradicts our derived equation where $(\frac{1-m}{1+m})\cot\phi = \tan\theta$. Hence, Option 2 is incorrect. - **Option 3: $2\tan\theta$**: This option implies that the expression is twice the value of $\tan\theta$, which does not match our derivation. Therefore, Option 3 is incorrect. - **Option 4: None of these**: Since we have shown that the expression equals $\tan\theta$, Option 4 is incorrect. In conclusion, the correct answer is indeed Option 1, $\tan\theta$, as derived from the given equation using trigonometric identities and logical reasoning.

Question

If $cos(	heta+\phi)=m~cos(	heta-\phi)$, then $(\frac{1-m}{1+m})cot~\phi$ is equal to:

Accepted Answer

To solve the problem, we start with the given equation:

$$ \cos(\theta + \phi) = m \cos(\theta - \phi) $$

We can use the sum and difference identities for cosine:

$$ \cos(\theta + \phi) = \cos\theta \cos\phi - \sin\theta \sin\phi $$
$$ \cos(\theta - \phi) = \cos\theta \cos\phi + \sin\theta \sin\phi $$

Substituting these into the given equation, we have:

$$ \cos\theta \cos\phi - \sin\theta \sin\phi = m (\cos\theta \cos\phi + \sin\theta \sin\phi) $$

Expanding the right side:

$$ \cos\theta \cos\phi - \sin\theta \sin\phi = m \cos\theta \cos\phi + m \sin\theta \sin\phi $$

Rearranging terms gives:

$$ \cos\theta \cos\phi - m \cos\theta \cos\phi = m \sin\theta \sin\phi + \sin\theta \sin\phi $$

Factor out common terms:

$$ (1 - m) \cos\theta \cos\phi = (1 + m) \sin\theta \sin\phi $$

Divide both sides by $\cos\theta \cos\phi$:

$$ \frac{1 - m}{1 + m} = \frac{\sin\theta \sin\phi}{\cos\theta \cos\phi} $$

This simplifies to:

$$ \frac{1 - m}{1 + m} = \tan\theta \tan\phi $$

Now, we need to find $(\frac{1-m}{1+m})\cot\phi$:

$$ (\frac{1-m}{1+m})\cot\phi = \tan\theta $$

Thus, the expression $(\frac{1-m}{1+m})\cot\phi$ is equal to $\tan\theta$, which corresponds to Option 1.

**Explanation for Other Options:**

- **Option 2: $-\tan\theta$**: This option suggests that the expression is negative, which contradicts our derived equation where $(\frac{1-m}{1+m})\cot\phi = \tan\theta$. Hence, Option 2 is incorrect.

- **Option 3: $2\tan\theta$**: This option implies that the expression is twice the value of $\tan\theta$, which does not match our derivation. Therefore, Option 3 is incorrect.

- **Option 4: None of these**: Since we have shown that the expression equals $\tan\theta$, Option 4 is incorrect.

In conclusion, the correct answer is indeed Option 1, $\tan\theta$, as derived from the given equation using trigonometric identities and logical reasoning.