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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.