Question

1-sec theta/tan theta+tan theta/1-sec theta=-2csc theta
Verify that each equation is an identity.

Answer 1

\[\huge \frac{1-\sec \theta}{\tan \theta}+ \frac{\tan \theta}{1-\sec \theta}=-2\csc \theta \]
Perhaps this is youer question you have posted above:
Now, \[\huge LHS= \frac{1-\sec \theta}{\tan \theta}+ \frac{\tan \theta}{1-\sec \theta}\]
\[\huge =\frac {(1-\sec \theta)^2 + \tan^2 \theta}{\tan \theta (1-\sec \theta)}\]
\[\huge =\frac {1+\sec^2 \theta -2\sec \theta + \sec^2 \theta-1}{\tan \theta (1-\sec \theta)}\]
\[\huge =\frac {2\sec^2 \theta -2\sec \theta}{\tan \theta (1-\sec \theta)}\]
\[\huge =\frac {2\sec \theta (\sec \theta -1)}{\tan \theta (1-\sec \theta)}\]
\[\huge =\frac {-2\sec \theta (1-\sec \theta)}{\tan \theta (1-\sec \theta)}\]
\[\huge =\frac {-2\sec \theta }{\tan \theta}\]
\[\huge =-2\frac { \frac{1}{\cos \theta} }{\frac{\sin \theta}{\cos \theta}}\]
\[\huge = \frac{-2}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\]
\[\huge = \frac{-2}{1} \times \frac{1}{\sin \theta}\]
\[\huge = -2 \times \csc \theta\]
\[\huge LHS = -2\csc \theta = RHS\]

@aliehslee
Hence the given equation is an identity;