Question

If $\tan \theta =\frac{a}{b}$,find the value of $\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$

Answer 1

$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{a}{b}$

$\Rightarrow \sin \theta =\frac{a\cos \theta }{b}$

Substituting in $\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$ we get

$=\frac{\cos \theta +\frac{a\cos \theta }{b}}{\cos \theta -\frac{a\cos \theta }{b}}$

$=\frac{\frac{b\cos \theta +a\cos \theta }{b}}{\frac{b\cos \theta -a\cos \theta }{b}}$

$=\frac{(b+a)\cos \theta }{(b-a)\cos \theta }$

$=\frac{a+b}{b-a}$