Question

Prove the following:
$\frac{1}{cosec\Theta -cot\Theta }-\frac{1}{sin\Theta }=\frac{1}{sin\Theta }-\frac{1}{cosec\Theta +cot\Theta }$

Answer 1

Let the $L.H.S$ of the given equation

$\begin{array}{c}LHS=\frac{1}{\cos ec\theta -cot\theta }-\frac{1}{\sin \theta }\\ =\frac{1}{cosec\theta -\cot \theta }\times \frac{\cos ec\theta +\cot \theta }{\cos ec\theta +\cot \theta }-\frac{1}{\sin \theta }\\ =\frac{\cos ec\theta +\cot \theta }{\cos e{c}^2\theta -{\cot }^2\theta }-\frac{1}{\sin \theta }\\ =\frac{\cos ec\theta +\cot \theta }{1}-\frac{1}{\sin \theta }\\ =\frac{\sin \theta \left(\cos ec\theta +\cot \theta \right)-1}{\sin \theta }\\ =\frac{1+\cos \theta -1}{\sin \theta }\\ =\frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta }\\ =\frac{1}{\sin \theta }+\cot \theta -\cos ec\theta \\ =\frac{1}{\sin \theta }-\left(\cos ec\theta -\cot \theta \right)\\ =\frac{1}{\sin \theta }-\left(\cos ec\theta -\cot \theta \right)\times \frac{\left(\cos ec\theta +\cot \theta \right)}{\left(\cos ec\theta +\cot \theta \right)}\\ =\frac{1}{\sin \theta }-\frac{1}{\left(\cos ec\theta +\cot \theta \right)}\end{array}$

$LHS=RHS$