Question

Prove that $(cosec\theta -\sec \theta )(\cot \theta -\tan \theta )=(cosec\theta +\sec \theta )(\sec \theta cosec\theta -2)$

Answer 1

LHS$=(cosec\theta -\sec \theta )(\cot \theta -\tan \theta ).$

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

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

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

$=\left(\frac{(1-2\cos \theta \sin \theta )(\cos \theta +\sin \theta )}{{\sin }^2\theta {\cos }^2\theta }\right)$

$=\left(\frac{(1-2\cos \theta \sin \theta )(\cos \theta +\sin \theta )}{\left(\sin \theta \cos \theta \right)\left(\sin \theta \cos \theta \right)}\right)$

$=(cosec\theta \sec \theta -2)(cosec\theta +\sec \theta )$

$=RHS$

Hence proved