Question

Solve $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }$

• A. $1+\tan \theta +\cot \theta$
• B. $1-\tan \theta +\cot \theta$
• C. $1+\tan \theta -\cot \theta$
• D. None of these

Answer 1

The correct option is A $1+\tan \theta +\cot \theta$

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

$=\frac{\frac{\sin \theta }{\cos \theta }}{1-\frac{\cos \theta }{\sin \theta }}+\frac{\frac{\cos \theta }{\sin \theta }}{1-\frac{\sin \theta }{\cos \theta }}$

$=\frac{{\sin }^2\theta }{\cos \theta \left(\sin \theta -\cos \theta \right)}+\frac{{\cos }^2\theta }{\sin \theta \left(\cos \theta -\sin \theta \right)}$

$=\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta \cos \theta \left(\sin \theta -\cos \theta \right)}$

$=\frac{\left(\sin \theta -\cos \theta \right)\left({\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta \right)}{\sin \theta \cos \theta \left(\sin \theta -\cos \theta \right)}$

$=\frac{{\sin }^2\theta +{\cos }^2\theta +\sin \theta \cos \theta }{\sin \theta \cos \theta }$

$=\frac{{\sin }^2\theta }{\sin \theta \cos \theta }+\frac{{\cos }^2\theta }{\sin \theta \cos \theta }+\frac{\sin \theta \cos \theta }{\sin \theta \cos \theta }$

$=1+\tan \theta +\cot \theta$