Question

Given \((\sin \theta - \csc \theta)^2 + (\cos \theta + \sec \theta)^2 = 5\) and \(\theta\) lies in the third quadrant, find \((\sin \theta + \cos \theta)^3\).

• A. \(-2\sqrt{2}\)
• B. \(2\sqrt{2}\)
• C. \(4\)
• D. \(-4\)

Hint:

Verify the trigonometric identities and ensure the quadrant-specific signs are accurately applied to obtain the correct cubic power results.

Answer 1

The Correct Option is A

First, simplify the given equation by expanding the terms: \[ (\sin \theta - \csc \theta)^2 + (\cos \theta + \sec \theta)^2 = 5 \] \[ (\sin^2 \theta - 2 \sin \theta \csc \theta + \csc^2 \theta) + (\cos^2 \theta + 2 \cos \theta \sec \theta + \sec^2 \theta) = 5 \] \[ (\sin^2 \theta - 2 + \frac{1}{\sin^2 \theta}) + (\cos^2 \theta + 2 + \frac{1}{\cos^2 \theta}) = 5 \] \[ \sin^2 \theta + \cos^2 \theta + \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = 5 \] Using \(\sin^2 \theta + \cos^2 \theta = 1\): \[ 1 + \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = 5 \] \[ \frac{1}{\sin^2 \theta} + \frac{1}{\cos^2 \theta} = 4 \] Now, focus on finding \((\sin \theta + \cos \theta)^3\). Start with expressing \(\sin \theta + \cos \theta\) using the angle-sum formula and Pythagorean identity: \[ \sin \theta + \cos \theta = \sqrt{2} \sin(\theta + \pi/4) \] In the third quadrant, both \(\sin \theta\) and \(\cos \theta\) are negative, and their sum squared would be: \[ (\sin \theta + \cos \theta)^2 = 2(\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta) \] \[ = 2(1 + \sin 2\theta) \] For the third quadrant where \(180^\circ<\theta<270^\circ\), \(\sin 2\theta\) is negative. Assuming \( \theta = 225^\circ \) for calculation: \[ \sin 225^\circ = -\frac{\sqrt{2}}{2} \] \[ (\sin \theta + \cos \theta)^2 = 2(1 - \frac{\sqrt{2}}{2}) \] \[ (\sin \theta + \cos \theta) = -\sqrt{2 - \sqrt{2}} \] Cubing this result: \[ (\sin \theta + \cos \theta)^3 = (-\sqrt{2 - \sqrt{2}})^3 = -2\sqrt{2} (since\ \theta\ in\ the\ third\ quadrant) \]