Question

Prove that $\frac{\cot \theta }{\csc \theta +1}+\frac{\csc \theta +1}{\cot \theta }=2\sec \theta$

Answer 1

$\frac{\cot \theta }{\csc \theta +1}+\frac{\csc \theta +1}{\cot \theta }=2\sec \theta$

Taking L.H.S., we have

$\Rightarrow \frac{\cot \theta }{\csc \theta +1}+\frac{\csc \theta +1}{\cot \theta }$

$\Rightarrow =\frac{{\cot }^2\theta +{(\csc \theta +1)}^2}{(\csc \theta +1)\cot \theta }$

$\Rightarrow =\frac{{\cot }^2\theta +({\csc }^2\theta +1+2\csc \theta )}{(\csc \theta +1)\cot \theta }$

$\Rightarrow =\frac{{\cot }^2\theta +1+{\csc }^2\theta +2\csc \theta }{(\csc \theta +1)\cot \theta }$

$\Rightarrow =\frac{{\csc }^2\theta +{\csc }^2\theta +2\csc \theta }{(\csc \theta +1)\cot \theta }(\because 1+{\cot }^2\theta ={\csc }^2\theta )$

$\Rightarrow =\frac{2{\csc }^2\theta +2\csc \theta }{(\csc \theta +1)\cot \theta }$

$\Rightarrow =\frac{2\csc \theta (\csc \theta +1)}{(\csc \theta +1)\cot \theta }$

$\Rightarrow =\frac{2\csc \theta }{\cot \theta }$

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

$\Rightarrow =\frac{2}{\cos \theta }$

$\Rightarrow =2\sec \theta$

$=R.H.S.$

Hence, it is proved that L.H.S. = R.H.S., i.e., $2\sec \theta$.