1. Prove the following identities: \( \begin{array}{ll}\text { a) }\left(1-\sin ^{2} \theta\right)\left(1+\tan ^{2} \theta\right)=1 & \text { b) }(\sin \theta+\cos \theta)^{2}=1+2 \sin \theta \cos \theta \\ \text { c) } \sin ^{4} x-\cos ^{4} x=\sin ^{2} x-\cos ^{2} x & \text { d) } \cos \theta\left(1+\tan ^{2} \theta\right)=\frac{1}{\cos \theta} \\ \text { e) } \frac{\tan ^{2} x}{1+\tan ^{2} x}=\sin ^{2} x & \text { f) } \sin x \tan x+\cos x=\frac{1}{\cos x} \\ \text { g) } \frac{\sin \theta \cos \theta}{\cos ^{2} \theta-\sin ^{2} \theta}=\frac{\tan \theta}{1-\tan ^{2} \theta} & \text { h) }(\sin x+\tan x)\left(\frac{\sin x}{1+\cos x}\right)=\sin x \tan x \\ \text { i) } \frac{\sin ^{2} 70^{\circ}+\sin ^{2} 20^{\circ}}{\sin ^{2} 40^{\circ}\left(1+\tan ^{2} 50^{\circ}\right)}=1 & \text { j) } \frac{\sin 70^{\circ} \cos 175^{\circ}}{\cos 340^{\circ} \cos 185^{\circ}}=1\end{array} \)

For part (a), we start with the left-hand side: \[ (1-\sin^2 \theta)(1+\tan^2 \theta) = \cos^2 \theta \cdot (1 + \frac{\sin^2 \theta}{\cos^2 \theta}) = \cos^2 \theta \cdot \frac{1}{\cos^2 \theta} = 1. \] Thus, the identity holds! Moving to part (b): \[ (\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2\sin \theta \cos \theta + \cos^2 \theta = 1 + 2 \sin \theta \cos \theta. \] This confirms the identity as well! The fun part of trigonometric identities is how they weave into our daily lives—think of how angles are involved in everything from engineering to architecture. Ever noticed how the tilt of a roof relates to sine and cosine? It’s literally a matter of staying dry! And if you want to dive deeper into trigonometric identities, grab a classic like "Trigonometry For Dummies." It’s filled with relatable examples and witty commentary, making complex concepts feel like a breeze. Plus, you can’t beat a little humor while trying to prove those tricky identities!