Question

Show that $\frac{\tan \theta }{(1-\cot \theta )}+\frac{\cot \theta }{(1-\tan \theta )}=(1+\sec \theta \text{cosec}\theta )$

Answer 1

LHS

$=\frac{\tan \theta }{(1-\cot \theta )}+\frac{\cot }{(1-\tan \theta )}$

[Using: $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and $\cot \theta =\frac{\cos \theta }{\sin \theta }$ ]

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

$=\frac{\sin \theta \tan \theta }{(\sin \theta -\cos \theta )}+\frac{\cos \theta \cot \theta }{(\cos \theta -\sin \theta )}$

$=\frac{\sin \theta \times \frac{\sin \theta }{\cos \theta }}{(\sin \theta -\cos \theta )}-\frac{\cos \theta \times \frac{\cos \theta }{\sin \theta }}{(\sin \theta -\cos \theta )}$

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

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

[Using identity: $(a^3-b^3)=(a-b)(a^2+ab+b^2)$

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

[Using identity:${\sin }^2\theta +{\cos }^2\theta =1$ ]

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

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

[Using identity: $\frac{1}{\sin \theta }=\text{cosec}\theta and\frac{1}{\cos \theta }=\sec \theta$ ]

$=\sec \theta \text{cosec}\theta +1$

RHS

$=1+\sec \theta \text{cosec}\theta$
Hence Proved.