Question

If \(\alpha + \beta = \frac{\pi}{2}\) and \(\beta + \gamma = \alpha\), then the value of \(\tan\alpha\) is:

• A. \(\tan\beta + \tan\gamma\)
• B. \(2(\tan\beta + \tan\gamma)\)
• C. \(\tan\beta + 2\tan\gamma\)
• D. \(2\tan\beta + \tan\gamma\)

Hint:

Use trigonometric identities for angle sums and double angles to simplify complex expressions involving tangents and cotangents.

Answer 1

The Correct Option is C

We are given the following conditions: \[ \alpha + \beta = \frac{\pi}{2}, \quad \beta + \gamma = \alpha. \] From the first condition, we can express \(\alpha\) as: \[ \alpha = \frac{\pi}{2} - \beta. \] Substitute this expression for \(\alpha\) in the second condition: \[ \beta + \gamma = \frac{\pi}{2} - \beta. \] Rearrange this to solve for \(\gamma\): \[ \gamma = \frac{\pi}{2} - 2\beta. \] Now, we need to find \(\tan\alpha\). Using the identity for \(\alpha = \frac{\pi}{2} - \beta\), we have: \[ \tan\alpha = \tan\left(\frac{\pi}{2} - \beta\right) = \cot\beta. \] Next, express \(\cot\beta\) in terms of \(\tan\beta\): \[ \cot\beta = \frac{1}{\tan\beta}. \] Now, from the expression for \(\gamma\), we know \(\gamma = \frac{\pi}{2} - 2\beta\). Therefore: \[ \tan\gamma = \tan\left(\frac{\pi}{2} - 2\beta\right) = \cot(2\beta). \] Using the double angle identity for cotangent: \[ \cot(2\beta) = \frac{1 - \tan^2\beta}{2\tan\beta}. \] Substitute this into the equation for \(\tan\alpha\): \[ \tan\alpha = \tan\beta + 2\tan\gamma = \tan\beta + 2 \cdot \frac{1 - \tan^2\beta}{2\tan\beta}. \] Simplify the equation: \[ \tan\alpha = \tan\beta + \frac{1 - \tan^2\beta}{\tan\beta}. \] Thus, the value of \(\tan\alpha\) is: \[ \tan\alpha = \tan\beta + 2\tan\gamma. \]