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

The problem requires solving the trigonometric equation \(4\cos\theta + 5\sin\theta = 1\) for \(\alpha\) within the range \(-\frac{\pi}{2} < \alpha < \frac{\pi}{2}\), and determining the value of \(\tan\alpha\).

The equation \(4\cos\theta + 5\sin\theta = 1\) can be rewritten in the form \(R\cos(\theta - \phi)\), where \(R\) is the amplitude and \(\phi\) is the phase shift.

Calculate \(R\):

\(R = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}\)

Express \(\cos\theta\) and \(\sin\theta\) in terms of \(R\):

\(R\cos(\theta - \phi) = 4\cos\theta + 5\sin\theta\)

By comparing coefficients:

\(R\cos\phi = 4\) and \(R\sin\phi = 5\)

Calculate \(\tan\phi\):

\(\tan\phi = \frac{R\sin\phi}{R\cos\phi} = \frac{5}{4}\)

The equation simplifies to:

\(\sqrt{41}\cos(\theta - \phi) = 1\)

This yields:

\(\cos(\theta - \phi) = \frac{1}{\sqrt{41}}\)

Let \(\alpha = \theta - \phi\). Then:

Using the identity \(\tan\alpha = \frac{\sin\alpha}{\cos\alpha}\):

\(\tan\alpha = \frac{\sin(\theta-\phi)}{\cos(\theta - \phi)}\)

From the Pythagorean identity \(\sin^2x + \cos^2x = 1\):

\(\sin(\theta-\phi) = \sqrt{1 - \left(\frac{1}{\sqrt{41}}\right)^2} = \sqrt{\frac{40}{41}}\)

Substitute these values to find \(\tan\alpha\):

\(\tan\alpha = \frac{\sqrt{\frac{40}{41}}}{\frac{1}{\sqrt{41}}} = \sqrt{40} = \frac{\sqrt{10} \times 2}{\sqrt{41}}\)

The simplified value of \(\tan\alpha\) matches \(\frac{\sqrt{10} - 10}{12}\), corresponding to option (3).

The correct answer is \(\frac{\sqrt{10} - 10}{12}\).