Question

If $\sec \theta +\tan \theta =4$, find $\sin \theta ,\cos \theta$.

Answer 1

$\sec \theta +\tan \theta =4$

i) $\sin \theta$

$\sec { \theta } +\tan { \theta } =4

therefore \cfrac { 1 }{ \cos { \theta } } +\cfrac { \sin { \theta } }{ \cos { \theta } } =4$

$\therefore \frac{1+\sin \theta }{\cos \theta }=4$

$\therefore {(1+\sin \theta )}^2={\left(4\cos \theta \right)}^2$

$\therefore {(1+\sin \theta )}^2=8{\cos }^2\theta$

$\therefore {(1+\sin \theta )}^2=8(1-{\sin }^2\theta )$

$\therefore {(1+\sin \theta )}^2=8(1-\sin \theta )(1+\sin \theta )$

$\therefore 1+\sin \theta =8(1-\sin \theta )$

$\therefore 1+\sin \theta =8-8\sin \theta$

$\therefore 9\sin \theta =7$

$\therefore \sin \theta =\frac{7}{9}$

ii) $\cos \theta$

$\sec \theta +\tan \theta =4$

$\therefore \frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=4$

$\therefore 1+\sin \theta =4\cos \theta$

$\therefore \therefore 1+\frac{7}{9}=4\cos \theta$

$\therefore \frac{16}{9}=4\cos \theta$

$\therefore \cos \theta =\frac{4}{9}$