Let's solve each question **step by step** with the **used formulas**: --- ### **19. Convert \( x^2 - y^2 = 3y \) to Polar Equation** **Formulas:** - \( x = r\cos\theta \) - \( y = r\sin\theta \) **Solution:** \[ x^2 - y^2 = 3y \] Substitute: \[ (r\cos\theta)^2 - (r\sin\theta)^2 = 3(r\sin\theta) \] \[ r^2(\cos^2\theta - \sin^2\theta) = 3r\sin\theta \] If \( r \neq 0 \), divide both sides by \( r \): \[ r(\cos^2\theta - \sin^2\theta) = 3\sin\theta \] \[ r = \frac{3\sin\theta}{\cos^2\theta - \sin^2\theta} \] --- ### **20. Convert \( \theta = \frac{\pi}{4} \) to Cartesian Equation** **Formula:** \( \tan\theta = \frac{y}{x} \) **Solution:** \[ \theta = \frac{\pi}{4} \] \[ \tan\theta = \tan\left(\frac{\pi}{4}\right) = 1 \] \[ \frac{y}{x} = 1 \implies y = x \] --- ### **21. Give two more polar coordinates for \( (-3, \pi) \)** **Formula:** \( (r, \theta) = (-r, \theta + \pi) \) **Solution:** - \( (-3, \pi) \) - \( (3, 0) \) (add \(\pi\) to angle, change sign of r) - \( (-3, 3\pi) \) (add \(2\pi\) to angle, r unchanged) --- ### **22. Find the exact value** #### (a) \( \sin^{-1}\left(-\frac{1}{2}\right) \) **Formula:** \( \sin^{-1}(x) \) - \( \sin^{-1}\left(-\frac{1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) \) - \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \) **Answer:** \( -\frac{\pi}{6} \) --- #### (b) \( \cos^{-1}\left(-\frac{1}{2}\right) \) **Formula:** \( \cos^{-1}(x) \) - \( \cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3} \) --- #### (c) \( \cos^{-1}\left(-\frac{\sqrt{2}}{2}\right) \) - \( \cos^{-1}\left(-\frac{\sqrt{2}}{2}\right) = \frac{3\pi}{4} \) --- #### (d) \( \tan^{-1}(-1) \) - \( \tan^{-1}(-1) = -\frac{\pi}{4} \) --- ### **23. \( \cos(\tan^{-1}(4) - \sin^{-1}(\frac{5}{9})) \)** Let \( A = \tan^{-1}(4) \), \( B = \sin^{-1}(\frac{5}{9}) \). **Formula:** \( \cos(A - B) = \cos A \cos B + \sin A \sin B \) - \( \tan A = 4 \rightarrow \sin A = \frac{4}{\sqrt{17}}, \cos A = \frac{1}{\sqrt{17}} \) - \( \sin B = \frac{5}{9} \rightarrow \cos B = \sqrt{1 - \left(\frac{5}{9}\right)^2} = \frac{8}{9} \) Now, \[ \cos(A - B) = \frac{1}{\sqrt{17}} \cdot \frac{8}{9} + \frac{4}{\sqrt{17}} \cdot \frac{5}{9} = \frac{8}{9\sqrt{17}} + \frac{20}{9\sqrt{17}} = \frac{28}{9\sqrt{17}} \] --- ### **24. If \( \csc 2\theta = -\frac{14}{9} \), \( -\pi \leq 2\theta \leq 0 \), find all six trig functions of \( \theta \)** - \( \csc 2\theta = \frac{1}{\sin 2\theta} \implies \sin 2\theta = -\frac{9}{14} \) - \( \cos^2 2\theta = 1 - \sin^2 2\theta = 1 - \left(-\frac{9}{14}\right)^2 = 1 - \frac{81}{196} = \frac{115}{196} \) - \( \cos 2\theta = \pm \sqrt{\frac{115}{196}} = \pm \frac{\sqrt{115}}{14} \) - Since \( -\pi \leq 2\theta \leq 0 \), \( 2\theta \) is in the third or fourth quadrant. Since \( \sin 2\theta \) is negative, and \( \csc \) is negative, this matches fourth quadrant, so \( \cos 2\theta > 0 \). - \( \cos 2\theta = \frac{\sqrt{115}}{14} \) - \( \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} = \frac{-\frac{9}{14}}{\frac{\sqrt{115}}{14}} = -\frac{9}{\sqrt{115}} \) - \( \theta = \frac{1}{2} \arcsin(-\frac{9}{14}) \) Now use half-angle identities for \( \theta \): - \( \sin^2\theta = \frac{1 - \cos 2\theta}{2} \) - \( \cos^2\theta = \frac{1 + \cos 2\theta}{2} \) - \( \sin\theta = -\sqrt{\frac{1 - \frac{\sqrt{115}}{14}}{2}} \) - \( \cos\theta = \sqrt{\frac{1 + \frac{\sqrt{115}}{14}}{2}} \) You can leave in this form or compute numerically if allowed. --- ### **25. Verify the identity \( \csc^2 t + \sec^2 t = \frac{\sec^2 t}{\sin^2 t} \)** - \( \csc t = \frac{1}{\sin t} \), \( \sec t = \frac{1}{\cos t} \) - \( \csc^2 t = \frac{1}{\sin^2 t} \), \( \sec^2 t = \frac{1}{\cos^2 t} \) Left side: \[ \csc^2 t + \sec^2 t = \frac{1}{\sin^2 t} + \frac{1}{\cos^2 t} = \frac{\cos^2 t + \sin^2 t}{\sin^2 t \cos^2 t} \] \[ \cos^2 t + \sin^2 t = 1 \] \[ \frac{1}{\sin^2 t \cos^2 t} \] Right side: \[ \frac{\sec^2 t}{\sin^2 t} = \frac{1/\cos^2 t}{\sin^2 t} = \frac{1}{\sin^2 t \cos^2 t} \] **Identity is verified.** --- If you need more steps for any specific question, let me know!