Question

If \( \tan(\theta - \phi) = \frac{3}{4} \) and \( \theta + \phi = \frac{\pi}{2} \), then the value of \[ \tan \left( \frac{\pi}{4} - \theta \right) + \tan \left( \frac{\pi}{4} - \phi \right) \] is:

• A. \( 1 \)
• B. \( 0 \)
• C. \( \frac{7}{4} \)
• D. \( \frac{5}{4} \)

Hint:

When dealing with trigonometric equations involving complementary angles (\( \theta + \phi = \frac{\pi}{2} \)), use identities like \( \tan \left( \frac{\pi}{2} - x \right) = \cot x \) and double-angle formulas to simplify expressions.

Answer 1

The Correct Option is B

We are given two conditions:

• \( \tan(\theta - \phi) = \frac{3}{4} \)
• \( \theta + \phi = \frac{\pi}{2} \)

From the second condition, we can express \( \phi \) in terms of \( \theta \): \[ \phi = \frac{\pi}{2} - \theta \] Now, substitute this into the expression we want to evaluate: \[ \tan \left( \frac{\pi}{4} - \theta \right) + \tan \left( \frac{\pi}{4} - \phi \right) = \tan \left( \frac{\pi}{4} - \theta \right) + \tan \left( \frac{\pi}{4} - \left( \frac{\pi}{2} - \theta \right) \right) \] \[ = \tan \left( \frac{\pi}{4} - \theta \right) + \tan \left( \frac{\pi}{4} - \frac{\pi}{2} + \theta \right) \] \[ = \tan \left( \frac{\pi}{4} - \theta \right) + \tan \left( \theta - \frac{\pi}{4} \right) \] Using the property \( \tan(-x) = -\tan(x) \), we have \( \tan \left( \theta - \frac{\pi}{4} \right) = - \tan \left( \frac{\pi}{4} - \theta \right) \). Therefore, the expression becomes: \[ \tan \left( \frac{\pi}{4} - \theta \right) - \tan \left( \frac{\pi}{4} - \theta \right) = 0 \] Since 0 is not among the options, let's re-examine the problem and our steps. From \( \theta + \phi = \frac{\pi}{2} \), we have \( \tan \phi = \tan(\frac{\pi}{2} - \theta) = \cot \theta = \frac{1}{\tan \theta} \). Using the first condition: \[ \tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1 + \tan \theta \tan \phi} = \frac{\tan \theta - \frac{1}{\tan \theta}}{1 + \tan \theta \cdot \frac{1}{\tan \theta}} = \frac{\frac{\tan^2 \theta - 1}{\tan \theta}}{2} = \frac{\tan^2 \theta - 1}{2 \tan \theta} = \frac{3}{4} \] \[ 4(\tan^2 \theta - 1) = 6 \tan \theta \] \[ 4 \tan^2 \theta - 6 \tan \theta - 4 = 0 \] \[ 2 \tan^2 \theta - 3 \tan \theta - 2 = 0 \] \[ (2 \tan \theta + 1)(\tan \theta - 2) = 0 \] So, \( \tan \theta = 2 \) or \( \tan \theta = -\frac{1}{2} \).
Case 1: \( \tan \theta = 2 \), then \( \tan \phi = \frac{1}{2} \). \[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan \theta}{1 + \tan \theta} = \frac{1 - 2}{1 + 2} = -\frac{1}{3} \] \[ \tan\left(\frac{\pi}{4} - \phi\right) = \frac{1 - \tan \phi}{1 + \tan \phi} = \frac{1 - \frac{1}{2}}{1 + \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} \] Sum = \( -\frac{1}{3} + \frac{1}{3} = 0 \).
Case 2: \( \tan \theta = -\frac{1}{2} \), then \( \tan \phi = -2 \). \[ \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - (-\frac{1}{2})}{1 + (-\frac{1}{2})} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3 \] \[ \tan\left(\frac{\pi}{4} - \phi\right) = \frac{1 - (-2)}{1 + (-2)} = \frac{3}{-1} = -3 \] Sum = \( 3 + (-3) = 0 \).