Question
If $\tan\theta=\frac{20}{21},$ show that $\frac{1-\sin\theta+\cos\theta}{1+\sin\theta+\cos\theta}=\frac{3}{7}.$
✓
Answer
$\tan\theta=\frac{20}{21}\text{ S.T }\frac{1-\sin\theta+\cos\theta}{1+\sin\theta+\cos\theta}=\frac{3}{7}$
$\tan\theta=\frac{\text{opposite side}}{\text{efficient side}}=\frac{20}{21}$
Let x be the hypotenuse By applying Pythagoras we get
$\text{AC}^2+\text{AB}^2+\text{BC}^2$
$\text{x}^2=(20)^2+(21)^2$
$\text{x}^2=400+441$
$\text{x}^2=841\Rightarrow\text{x}=29$
$\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{20}{29}$
$\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{21}{29}$
Substitute $\sin\theta,\cos\theta$ in equation we get
$\Rightarrow\frac{1-\sin\theta+\cos\theta}{1+\sin\theta+\cos\theta}$
$\Rightarrow\frac{1-\frac{20}{29}+\frac{21}{29}}{1+\frac{20}{29}+\frac{21}{29}}=\frac{\frac{29-20+21}{29}}{\frac{29+20+21}{29}}=\frac{30}{70}=\frac{3}{7}$
$\tan\theta=\frac{\text{opposite side}}{\text{efficient side}}=\frac{20}{21}$
Let x be the hypotenuse By applying Pythagoras we get
$\text{AC}^2+\text{AB}^2+\text{BC}^2$
$\text{x}^2=(20)^2+(21)^2$
$\text{x}^2=400+441$
$\text{x}^2=841\Rightarrow\text{x}=29$
$\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{20}{29}$
$\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{21}{29}$
Substitute $\sin\theta,\cos\theta$ in equation we get
$\Rightarrow\frac{1-\sin\theta+\cos\theta}{1+\sin\theta+\cos\theta}$
$\Rightarrow\frac{1-\frac{20}{29}+\frac{21}{29}}{1+\frac{20}{29}+\frac{21}{29}}=\frac{\frac{29-20+21}{29}}{\frac{29+20+21}{29}}=\frac{30}{70}=\frac{3}{7}$
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