If $tan\ θ = \frac{a}{b}$ , find the value of $\frac{cos\ θ+sin\ θ}{cos\ θ−sin\ θ}$.

Given:

$tan\ θ = \frac{a}{b}$.

To do:

We have to find the value of $\frac{cos\ θ+sin\ θ}{cos\ θ−sin\ θ}$.

Solution:  

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

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,

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

$\Rightarrow AC^2=(b)^2+(a)^2$

$\Rightarrow AC^2=b^2+a^2$

$\Rightarrow AC=\sqrt{a^2+b^2}$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{a}{\sqrt{a^2+b^2}}$

$cos\ \theta=\frac{AB}{AC}=\frac{b}{\sqrt{a^2+b^2}}$

 $\frac{cos\ θ+sin\ θ}{cos\ θ−sin\ θ}= \frac{\frac{b}{\sqrt{a^{2} +b^{2}}} +\frac{a}{\sqrt{a^{2} +b^{2}}}}{\frac{b}{\sqrt{a^{2} +b^{2}}} -\frac{a}{\sqrt{a^{2} +b^{2}}}}$

$=\frac{\frac{b+a}{\sqrt{a^{2} +b^{2}}}}{\frac{b-a}{\sqrt{a^{2} +b^{2}}}}$

$=\frac{a+b}{b-a}$

The value of $\frac{cos\ θ+sin\ θ}{cos\ θ−sin\ θ}$ is $\frac{a+b}{b-a}$.

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Updated on: 10-Oct-2022

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