Question

9.
Prove that
sin x -sin 3x/
sin^2x-cos^2x =
2 sin x

Answer 1

$\mathfrak{ \huge \underline{To~prove:}} \\\\\sf \frac{sin x -sin 3x} {sin^2x-cos^2x} =2 sin x$

$\mathfrak{ \huge \underline{Answer:}}$

By transformation formulae...

$\sf \frac{sin x -sin 3x} {sin^2x-cos^2x} \\\\\sf \longrightarrow \frac{2cosx(\frac{x+3x}{2}) .sin(\frac{x-3x}{2}) } {\frac{1-cos2x}{2}-\frac{1+cos2x}{2}} \\\\\sf \longrightarrow \frac {2cos2x.sin(-x)}{\frac {1-cos2x-1-cos2x} {2}} \\\\\sf \longrightarrow \frac {-2cos2x.sinx }{\frac {-cos2x-cos2x} {2}} \\\\\sf \longrightarrow \frac {-2cos2x.sinx }{\frac {-2cos2x} {2}} \\\\\sf \longrightarrow \frac {-2cos2x.sinx }{\frac {-\cancel{2} cos2x} {\cancel{2}}} \\\\\sf \longrightarrow \frac{-2cos2x.sinx} {-cos2x} \\\\\sf \longrightarrow \huge \boxed {2sinx}$

Hence, proved