Question

If \(\cos\theta=\frac{5}{11}\) and \(\tan\theta\lt0\), then the value of \(\sin\theta\) is equal to

• A. \(\frac{8\sqrt6}{11}\)
• B. \(\frac{-8\sqrt6}{11}\)
• C. \(\frac{4\sqrt6}{11}\)
• D. \(\frac{-4\sqrt6}{11}\)
• E. \(\frac{6}{11}\)

Answer 1

The Correct Option is D

We are given that: \[ \cos \theta = \frac{5}{11} \quad \text{and} \quad \tan \theta < 0. \] To find \( \sin \theta \), we will use the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1. \] Step 1: Substitute the given value of \( \cos \theta \) We know that \( \cos \theta = \frac{5}{11} \), so: \[ \sin^2 \theta + \left( \frac{5}{11} \right)^2 = 1 \] \[ \sin^2 \theta + \frac{25}{121} = 1. \] Step 2: Solve for \( \sin^2 \theta \) Subtract \( \frac{25}{121} \) from both sides: \[ \sin^2 \theta = 1 - \frac{25}{121} = \frac{121}{121} - \frac{25}{121} = \frac{96}{121}. \] Step 3: Take the square root of both sides \[ \sin \theta = \pm \sqrt{\frac{96}{121}} = \pm \frac{\sqrt{96}}{11}. \] Since \( \sqrt{96} = 4 \sqrt{6} \), we have: \[ \sin \theta = \pm \frac{4 \sqrt{6}}{11}. \] Step 4: Use the condition \( \tan \theta < 0 \) Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \tan \theta < 0 \), we must choose the negative value for \( \sin \theta \) because \( \cos \theta = \frac{5}{11} > 0 \). Thus, \[ \sin \theta = \frac{-4 \sqrt{6}}{11}. \]

The correct option is (D) : \(\frac{-4\sqrt6}{11}\)

Answer 2

We are given that \(\cos\theta = \frac{5}{11}\) and \(\tan\theta \lt 0\).

We know that \(\sin^2\theta + \cos^2\theta = 1\). Therefore, \(\sin^2\theta = 1 - \cos^2\theta\).

Substituting the given value of \(\cos\theta\), we have:

\(\sin^2\theta = 1 - \left(\frac{5}{11}\right)^2 = 1 - \frac{25}{121} = \frac{121 - 25}{121} = \frac{96}{121}\)

Taking the square root, we get:

\(\sin\theta = \pm\sqrt{\frac{96}{121}} = \pm\frac{\sqrt{96}}{11}\)

Since \(96 = 16 \cdot 6\), we have \(\sqrt{96} = \sqrt{16 \cdot 6} = 4\sqrt{6}\). Therefore:

\(\sin\theta = \pm\frac{4\sqrt{6}}{11}\)

We are given that \(\tan\theta \lt 0\). Since \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), and we know that \(\cos\theta = \frac{5}{11} \gt 0\), it must be the case that \(\sin\theta \lt 0\) for \(\tan\theta\) to be negative.

Therefore, we choose the negative root:

\(\sin\theta = -\frac{4\sqrt{6}}{11}\)