COMPLETE THESE ACTIVITIES.
1.1 Given the point \( T(3,4) \), find \( \sin \theta \) and \( \csc \theta \) and show that their product is 1 .
1.2 Given the point \( T(-3,-4) \), find \( \tan \theta \) and \( \cot \theta \) and show that their product is 1 .
1.3 Given \( \theta=\frac{\pi}{4} \), show that \( \cos \theta \sec \theta=1 \).
1.4 Given \( \theta=\frac{\pi}{3} \), show that \( \sin \theta \csc \theta=1 \).
1.5 Simplify \( \tan \theta \sin \theta \cot \theta \).
1.6 Simplify \( \frac{1}{\cos \theta \sec \theta} \cdot \cot \theta \)
1.7 Simplify \( 1+\tan \theta \cot \theta-\frac{\sin \theta \csc \theta}{2} \).
1.8 Simp11fy \( \frac{\sin \theta}{\csc \theta} \) by using \( \csc \theta=\frac{1}{\sin \theta} ; \theta \neq 0 \).
1.9 Simplify \( \frac{\tan \theta}{\cot \theta} \)
1.10 Simplify \( \frac{\sec \theta}{\cos \theta} \)

Step 1 ：1.1 \(\sin\theta = \frac{4}{5}, \csc\theta = \frac{5}{4}, \sin\theta\cdot\csc\theta = \frac{4}{5} \cdot \frac{5}{4} = 1\)

Step 2 ：1.2 \(\tan\theta = -\frac{4}{3}, \cot\theta = -\frac{3}{4}, \tan\theta\cdot\cot\theta = -\frac{4}{3} \cdot -\frac{3}{4} = 1\)

Step 3 ：1.3 \(\cos\theta = \frac{1}{\sqrt{2}}, \sec\theta = \sqrt{2}, \cos\theta\cdot\sec\theta = \frac{1}{\sqrt{2}} \cdot \sqrt{2} = 1\)

Step 4 ：1.4 \(\sin\theta = \frac{\sqrt{3}}{2}, \csc\theta = \frac{2}{\sqrt{3}}, \sin\theta\cdot\csc\theta = \frac{\sqrt{3}}{2} \cdot \frac{2}{\sqrt{3}} = 1\)

Step 5 ：1.5 \(\tan\theta\sin\theta\cot\theta = \tan\theta\cdot (1)(\cot\theta) = \tan\theta\cdot\cot\theta = 1\)

Step 6 ：1.6 \(\frac{1}{\cos\theta\sec\theta} \cdot \cot\theta = \frac{\cot\theta}{\cos\theta\sec\theta} = \cot\theta \div 1 = \cot\theta\)

Step 7 ：1.7 \(1+\tan\theta\cot\theta-\frac{\sin\theta\csc\theta}{2} = 1 + 1 - \frac{1}{2} = \frac{3}{2}\)

Step 8 ：1.8 \(\frac{\sin\theta}{\csc\theta} = \frac{\sin\theta}{\frac{1}{\sin\theta}} = \frac{\sin\theta}{1}(\sin\theta) = \sin^2\theta\)

Step 9 ：1.9 \(\frac{\tan\theta}{\cot\theta} = \frac{\tan\theta}{\frac{1}{\tan\theta}} = \tan^2\theta\)

Step 10 ：1.10 \(\frac{\sec\theta}{\cos\theta} = \frac{\sec\theta}{\frac{1}{\sec\theta}} = \sec^2\theta\)