Question

Prove:

$\left(\cos ec\theta -\sin \theta \right)\left(\sec \theta -\cos \theta \right)=\frac{1}{\tan \theta +\cot \theta }$

Answer 1

Consider the $LHS$

$\begin{array}{c}\left(\cos ec\theta -\sin \theta \right)\left(\sec \theta -\cos \theta \right)\\ =\left(\frac{1}{\sin \theta }-\sin \theta \right)\left(\frac{1}{\cos \theta }-\cos \theta \right)\\ =\left(\frac{1-{\sin }^2\theta }{\sin \theta }\right)\left(\frac{1-{\cos }^2\theta }{\cos \theta }\right)\\ =\frac{{\cos }^2\theta }{\sin \theta }\times \frac{{\sin }^2\theta }{\cos \theta }\\ =\sin \theta \cos \theta \end{array}$

now, Consider the $RHS$

$\begin{array}{c}\frac{1}{\tan \theta +\cot \theta }\\ =\frac{1}{\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }}\\ =\frac{1}{\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }}\\ =\frac{\sin \theta \cos \theta }{{\cos }^2\theta +{\sin }^2\theta }\\ =\sin \theta \cos \theta \left({\because \sin }^2\theta +{\cos }^2\theta =1\right)\end{array}$

Since, $LHS=RHS$

Therefore, $\left(\cos ec\theta -\sin \theta \right)\left(\sec \theta -\cos \theta \right)=\frac{1}{\tan \theta +\cot \theta }$