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

Given:

\( \tan \theta=\frac{20}{21} \)

To do:

We have to show that \( \frac{1-\sin \theta+\cos \theta}{1+\sin \theta+\cos \theta}=\frac{3}{7} \).

Solution:  

Let, in a triangle $ABC$ right-angled at $B$ and $\ tan\ \theta = tan\ A=\frac{20}{21}$.

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

Here,

$AC^2=AB^2+BC^2$

$\Rightarrow AC^2=(21)^2+(20)^2$

$\Rightarrow AC^2=441+400$

$\Rightarrow AC=\sqrt{841}=29$

Therefore,

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

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

Let us consider LHS,

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

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

$=\frac{50-20}{70}$

$=\frac{30}{70}$

$=\frac{3}{7}$

$=$ RHS

Hence proved.

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Updated on: 10-Oct-2022

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