Question

If $\cot \theta +\tan \theta =x$ and $\sec \theta -\cos \theta =y$, then show that $\sin \theta \cdot \cos \theta =\frac{1}{x}$ or $\sin \theta \cdot \tan \theta =y$ .

Answer 1

$\cot \theta +\tan \theta =x\sec \theta -\cos \theta =y$

$\left(i\right)\frac{1}{x}=\frac{1}{\cot \theta +\tan \theta }=\frac{1}{\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }}=\frac{\sin \theta \cos \theta }{{\sin }^2\theta +{\cos }^2\theta }\therefore \frac{1}{x}=\sin \theta \cos \theta$

$\left(ii\right)y=\sec \theta -\cos \theta y=\frac{1}{\cos \theta }-\cos \theta y=\frac{1-{\cos }^2\theta }{\cos \theta }=\frac{{\sin }^2\theta }{\cos \theta }=\tan \theta \sin \theta$