Question

can someone please help me with part 3 in the picture?? It's urgent

Answer 1

Answer:

Hope this helps.

$\displaystyle\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}\\ \\=\frac{\frac{\sin\theta}{\cos\theta}}{1-\frac{\cos\theta}{\sin\theta}}+\frac{\frac{\cos\theta}{\sin\theta}}{1-\frac{\sin\theta}{\cos\theta}}\\ \\=\frac{\sin^2\theta}{\cos\theta\sin\theta-\cos^2\theta}+\frac{\cos^2\theta}{\cos\theta\sin\theta-\sin^2\theta}\\ \\=\frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)}-\frac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}$

$\displaystyle=\frac{\sin^3\theta-\cos^3\theta}{\cos\theta\sin\theta(\sin\theta-\cos\theta)}\\ \\=\frac{(\sin\theta-\cos\theta)(\sin^2\theta+\cos\theta\sin\theta+\cos^2\theta)}{\cos\theta\sin\theta(\sin\theta-\cos\theta)}\\ \\=\frac{\cos\theta\sin\theta+1}{\cos\theta\sin\theta}\\ \\=1 + \frac{1}{\cos\theta\sin\theta}\\ \\=1 + \sec\theta\csc\theta$