Question

Prove the identity $(\text{cosec}\theta -\sin \theta )(\sec \theta -\cos \theta )=\frac{1}{\tan \theta +\cot \theta }$.

Answer 1

Now,
$(cosec\theta -\sin \theta )(\sec \theta -\cos \theta )$

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

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

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

Next, consider $\frac{1}{\tan \theta +\cot \theta }$

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

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

$=\sin \theta \cos \theta$ (2)

From (1) and (2), we get

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

Note

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

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

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

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

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