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,

$tanA=\frac{20}{21}$

We know that,

$tanA=\frac{oppositeSide}{adjacentSide}$

From Pythagoras theorem,

$(Hypotenuse{)}^2=(oppositeSide{)}^2+(adjacentSide{)}^2$

$(Hypotenuse{)}^2={20}^2+{21}^2$

$(Hypotenuse{)}^2=400+441=841$

$\left(Hypotenuse\right)=29$

$cosA=\frac{AdjacentSide}{Hypotenuse}=\frac{21}{29}$

$sinA=\frac{OppositeSide}{AdjacentSide}=\frac{20}{29}$

Therefore,

L..H.S

=$\frac{1-\sin \theta +\cos \theta }{1+\sin \theta +\cos \theta }$

=$\frac{1-\left(\frac{20}{29}\right)+\left(\frac{21}{29}\right)}{1+\left(\frac{20}{29}\right)+\left(\frac{21}{29}\right)}$

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

=$\frac{30}{70}$

=$\frac{3}{7}$=R.H.S

$\therefore \frac{1-\sin \theta +\cos \theta }{1+\sin \theta +\cos \theta }=\frac{3}{7}$

Hence proved