Question

The value of $\frac{\tan \alpha }{1-\cot \alpha }+\frac{\cot \alpha }{1-\tan \alpha }$ is identically equal to

• A. $\sec \alpha .\csc \alpha$
• B. $\sin \alpha .\cos \alpha$
• C. $\sec \alpha .\csc \alpha +1$
• D. $\sin \alpha .\cos \alpha +1$

Answer 1

Answer: (C)

$\sec \alpha .\csc \alpha +1$

$\frac{\tan \alpha }{1-\cot \alpha }+\frac{\cot \alpha }{1-\tan \alpha }$

$=\frac{\tan \alpha }{1-\frac{1}{\tan \alpha }}+\frac{\cot \alpha }{1-\tan \alpha }$

$=\frac{{\tan }^2\alpha }{\tan \alpha -1}-\frac{\cot \alpha }{\tan \alpha -1}$

$=\frac{{\tan }^2-\frac{1}{\tan \alpha }}{\tan \alpha -1}$

$=\frac{{\tan }^3\alpha -1}{\left(\tan \alpha \right)(\tan \alpha -1)}$

$=\frac{(\tan \alpha -1)({\tan }^2\alpha +\tan \alpha +1)}{\left(\tan \alpha \right)(\tan \alpha -1)}$

$=\frac{{\tan }^2\alpha +\tan \alpha +1}{\tan \alpha }$

$=\tan \alpha +1+\cot \alpha$

$=\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }+1$

$=\frac{{\sin }^2\alpha +{\cos }^2\alpha }{\cos \alpha \sin \alpha }+1$

$=\frac{1}{\cos \alpha \sin \alpha }+1$

$=\sec \alpha \csc \alpha +1$

Hence answer is C