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

Let us consider a right $\Delta ABC$ right angled at B

Now it is given that $\tan \theta =\frac{AB}{BC}=\frac{20}{21}$

So, if AB = 20k, then BC = 21k where k is a positive number

Using Pythagoras theorem, we have

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=(20k{)}^2+(21k{)}^2$

$A{C}^2=841k^2$

$AC=\sqrt{841k^2}=29k$

Now

$sin\theta =\frac{AB}{AC}=\frac{20}{29}$

$cos\theta =\frac{BC}{AC}=\frac{21}{29}$

Substitute these values in

$\frac{(1-sin\theta +cos\theta )}{(1+sin\theta +cos\theta )}$

$=\frac{1-\frac{20}{29}+\frac{21}{29}}{1+\frac{20}{29}+\frac{21}{29}}$

$=\frac{\frac{29-20+21}{29}}{\frac{29+20+21}{29}}$

$=\frac{30}{70}=\frac{3}{7}$