If \[ \tan x=\frac{a}{b} \] Show that \[ \frac{a\sin x-b\cos x} {a\sin x+b\cos x} = \frac{a^2-b^2}{a^2+b^2} \]

Solution:

Since \[ \tan x=\frac{a}{b} \]

\[ \sin x=\frac{a}{\sqrt{a^2+b^2}} \] and \[ \cos x=\frac{b}{\sqrt{a^2+b^2}} \]

Now,

\[ \frac{a\sin x-b\cos x} {a\sin x+b\cos x} \]

\[ = \frac{ a\left(\frac{a}{\sqrt{a^2+b^2}}\right) – b\left(\frac{b}{\sqrt{a^2+b^2}}\right) } { a\left(\frac{a}{\sqrt{a^2+b^2}}\right) + b\left(\frac{b}{\sqrt{a^2+b^2}}\right) } \]

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

\[ = \frac{a^2-b^2}{a^2+b^2} \]

Hence proved.