Question

Prove that:
${(\csc \theta +\cot \theta )}^2=\frac{\sec \theta +1}{\sec \theta -1}$

Answer 1

We need to prove ${(\csc \theta +\cot \theta )}^2=\frac{\sec \theta +1}{\sec \theta -1}$
LHS $={(\csc \theta +\cot \theta )}^2$
$={[\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }]}^2$
$={\left[\frac{1+\cos \theta }{\sin \theta }\right]}^2={\left[\frac{2{\cos }^2\theta /2}{2\sin \theta /2\cos \theta /2}\right]}^2[\because \sin 2\theta =\sin \theta \cos \theta ;1+\cos 2\theta =2{\cos }^2\theta ]$
$={\left[\frac{\cos \theta /2}{\sin \theta /2}\right]}^2$

LHS $={\cot }^2\theta /2$
Now, RHS $=\frac{\sec \theta +1}{\sec \theta -1}=\frac{1+\cos \theta }{1-\cos \theta }$
$=\frac{2{\cos }^2\theta /2}{2{\sin }^2\theta /2}[\because 1-\cos 2\theta =2{\sin }^2\theta ]$
$={\cot }^2\theta /2=$ LHS
$\therefore$ LHS $=$ RHS
Hence, proved.