Question

$\sin \theta (1+\tan \theta )+\cos \theta (1+\cot \theta )=sec\theta +cosec\theta$.

Answer 1

Let the problem be

$\begin{array}{c}LHS=\sin \theta \left(1+\tan \theta \right)+\cos \theta \left(1+\cot \theta \right)\\ =\sin \theta +\sin \theta \times \frac{\sin \theta }{\cos \theta }+\cos \theta +\cos \theta \times \frac{\cos \theta }{\sin \theta }\\ =\frac{\cos \theta {\sin }^2\theta +{\sin }^3\theta +{\cos }^2\theta \sin \theta +{\cos }^3\theta }{\cos \theta \sin \theta }\\ =\frac{\left(\sin \theta +\cos \theta \right)\left({\sin }^2\theta -\sin \theta \cos \theta +{\cos }^2\theta \right)+\sin \theta \cos \theta \left(\sin \theta +\cos \theta \right)}{\cos \theta \sin \theta }\\ =\frac{\left(\sin \theta +\cos \theta \right)\left({\sin }^2\theta +{\cos }^2\theta -\sin \theta \cos \theta +\sin \theta \cos \theta \right)}{\cos \theta \sin \theta }\\ =\frac{\left(\sin \theta +\cos \theta \right)\left(1\right)}{\cos \theta \sin \theta }\\ =\frac{\sin \theta }{\cos \theta \sin \theta }+\frac{\cos \theta }{\cos \theta \sin \theta }\\ =\frac{1}{\cos \theta }+\frac{1}{\sin \theta }\\ =\sec \theta +\cos ec\theta \\ =RHS\end{array}$