Question

Show that $(\csc \theta -\cot \theta {)}^2=\frac{1-\cos \theta }{1+\cos \theta }$

Answer 1

To prove $(cosec\theta -\cot \theta {)}^2=\frac{1-\cos \theta }{1+\cos \theta }$

$L.H.S(cosec\theta -\cot \theta {)}^2$

$={(\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta })}^2[cosecx=\frac{1}{\sin x},\cot x=\frac{\cos x}{\sin x}]$

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

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

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

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

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

$=R.H.S$