The value of $$\frac{\sin(90^o - \theta)\cos(90^o - \theta)\cot(90^o - \theta)}{\cos^2 \theta -1}$$​ is ___________.

Correct option is C

Given:

​$$\frac{\sin(90^o - \theta)\cos(90^o - \theta)\cot(90^o - \theta)}{\cos^2 \theta -1}$$​​

Formula Used:

​$$\tan \theta = \frac{\sin \theta}{\cos\theta}$$​​

​$$\sin^2 \theta + \cos^2 \theta = 1$$​​

​$$\sin(90^o - \theta) = \cos \theta$$​​

​$$\cos(90^o - \theta) = \sin \theta$$​​

​$$\cot(90^o - \theta) = \tan \theta$$​​

Solution:

​$$\frac{\sin(90^o - \theta)\cos(90^o - \theta)\cot(90^o - \theta)}{\cos^2 \theta -1}$$​​

$$\frac{\sin\theta\cos\theta\tan\theta}{\cos^2 \theta -1}$$​

​$$\frac{\sin \theta \sin \theta }{\cos^2 \theta -1}$$​​

​$$\frac{\sin^2 \theta}{\cos^2 \theta - 1}$$​​

​$$\frac{1 - \cos^2 \theta}{\cos^2 \theta - 1}$$​​

​$$= \frac{ -(\cos^2 \theta - 1)}{\cos^2 \theta - 1}$$​​

= -1