Question

If\(\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}\)is the solution of\( 4cos\theta+ 5sin\theta=1\)then the value of \(tan\alpha\) is

• A. \(\frac{10-\sqrt10}{6}\)
• B. \(\frac{10-\sqrt10}{12}\)
• C. \(\frac{\sqrt10-10}{12}\)
• D. \(\frac{\sqrt10-10}{6}\)

Answer 1

The Correct Option is C

Step 1. Rewrite the Equation in Terms of tan θ

Given 4 + 5 \(\tan θ\) = \(\sec θ\).

Step 2. Square Both Sides

Squaring both sides to eliminate \(\sec θ\), we get:

\(24 \tan^2 θ + 40 \tan θ + 15 = 0\)

Step 3. Solve for tan θ

Solving this quadratic equation, we find:

\(\tan θ = \frac{-10 \pm \sqrt{10}}{12}\)

Step 4. Choose the Correct Value Based on Range

Since \(-\frac{\pi}{2} < α < \frac{\pi}{2}\), we reject \(\tan α = \frac{-10 + \sqrt{10}}{12}\) and select:

\(\tan α = \frac{\sqrt{10} - 10}{12}\)

So, the correct answer is: \(\frac{\sqrt{10} - 10}{12}\)