Question

If $\tan \theta =\frac{20}{21}$, show that $\frac{1-\sin \theta +\cos \theta }{1+\sin \theta +\cos \theta }=\frac{3}{7}$

Answer 1

Given that $\tan \theta =\frac{20}{21}$. Let us consider the right triangle $ABC$ (see fig), with $AB=21$ and $BC=20$. By Pythagoras theorem, we have
$A{C}^2=A{B}^2+B{C}^2={20}^2+{21}^2=400+441=841$.
$\therefore AC=\sqrt{841}=29$.
$\sin \theta =\frac{BC}{AC}=\frac{20}{29},\cos \theta =\frac{AB}{AC}=\frac{21}{29}$
$1-\sin \theta +\cos \theta =1-\frac{20}{29}+\frac{21}{29}=\frac{29-20+21}{29}=\frac{30}{29}$
$1+\sin \theta +\cos \theta =1+\frac{20}{29}+\frac{21}{29}=\frac{29+20+21}{29}=\frac{70}{29}$
$\frac{1-\sin \theta +\cos \theta }{1+\sin \theta +\cos \theta }=\frac{\frac{30}{29}}{\frac{70}{29}}=\frac{30}{29}\times \frac{29}{70}=\frac{30}{70}=\frac{3}{7}$