Question

Proe that (cosec theta - sin sin theta) (sec theta - cos theta) = 1/tan theta + cot theta

Answer 1

Here $LHS=(\csc \theta-\sin \theta)(\sec \theta-\cos \theta)$.

We know,

$\sec \theta = \frac{1}{\cos \theta} \\ \csc \theta = \frac{1}{\sin \theta} \\ \sin^2 \theta+\cos^2 \theta=1$

Using the above identities,

$LHS=(\frac{1}{\sin \theta}-\sin \theta)(\frac{1}{\cos \theta}-\cos \theta) \\ LHS=\frac{(1- \sin^2 \theta)(1- \cos ^2 \theta)}{\sin \theta\cos \theta} \\ LHS=\frac{\sin^2 \theta \cos ^2 \theta}{\sin \theta\cos \theta} \\ LHS=\sin \theta\cos \theta$

$LHS=\sin \theta\cos \theta\\ LHS=\frac{\sin \theta \cos \theta}{1} \\ LHS=\frac{\sin \theta \cos \theta}{\sin^2 \theta+\cos^2 \theta} \\ LHS=\frac{1}{\frac{\sin^2 \theta }{\sin \theta \cos \theta} +\frac{\cos^2 \theta}{\sin \theta \cos \theta}} \\ LHS=\frac{1}{\tan \theta+\cot \theta}=RHS$