Question

Show that $\cot \theta +\tan \theta =\sec \theta \csc \theta$

Answer 1

$\cot \theta +\tan \theta =\sec \theta \csc \theta$

Taking L.H.S., we have

$\cot \theta +\tan \theta$

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

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

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

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

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

$\Rightarrow =\csc \theta .\sec \theta$

$=R.H.S.$

Hence, it is proved that L.H.S. = R.H.S., i.e., $\cot \theta +\tan \theta =\sec \theta \csc \theta$