# If $tan \theta = \frac{a}{b}$, prove that$\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$.

Given:

$tan\ θ = \frac{a}{b}$.

To do:

We have to prove that

$\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$.

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,

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

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

This implies,

Let us consider LHS,

$\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a\left(\frac{a}{\sqrt{a^{2} +b^{2}}}\right) -b\left(\frac{b}{\sqrt{a^{2} +b^{2}}}\right)}{a\left(\frac{a}{\sqrt{a^{2} +b^{2}}}\right) +b\left(\frac{b}{\sqrt{a^{2} +b^{2}}}\right)}$

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

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

$=$ RHS

Hence proved.

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Updated on: 10-Oct-2022

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