Question

Prove that $\sin \theta (1+\tan \theta )+\cos \theta (1+\cot \theta )=\sec \theta +\csc \theta$.

Answer 1

$\sin \theta (1+\tan \theta )+\cos \theta (1+\cot \theta )$

$=\sin \theta +\sin \theta \times \frac{\sin \theta }{\cos \theta }+\cos \theta +\cos \theta \times \frac{\cos \theta }{\sin \theta }$

$=(\sin \theta +\sin \theta )+\frac{{\sin }^2\theta }{\cos \theta }+\frac{{\cos }^2\theta }{\sin \theta }$

$=(\sin \theta +\cos \theta )+\frac{{\sin }^3\theta +{\cos }^3\theta }{\cos \theta \sin \theta }$

$=\frac{\cos \theta \sin \theta (\sin \theta +\cos \theta )+{\sin }^3\theta +{\cos }^3\theta }{\cos \theta \sin \theta }$

$=\frac{({\sin }^2\theta +{\cos }^2\theta )(\sin \theta +\cos \theta )}{\cos \theta \sin \theta }$

$=\frac{\sin \theta +\cos \theta }{\cos \theta \sin \theta }$

$=\frac{1}{\sin \theta }+\frac{1}{\cos \theta }$

$=\csc \theta +\sec \theta$