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Answer 1

To prove ---->

$\dfrac{sinx - sin3x}{ {sin}^{2}x - {cos}^{2}x } \: = 2 \: sinx$

Concept used ---->

1)

$sin3x = 3 \: sinx \: - \: 4 \: {sin}^{3} x$

2)

${cos}^{2} x \: = 1 \: - {sin}^{2} x$

Proof ---> LHS

$= \dfrac{sinx \: - \: sin3x}{ {sin}^{2} x - {cos}^{2}x }$

$= \dfrac{sinx \: - (3 \: sinx \: - 4 {sin}^{3} x)}{ {sin}^{2}x \: - ( \: 1 \: - {sin}^{2}x \: ) }$

$= \dfrac{sinx \: - 3sinx \: + 4 {sin}^{3}x }{ {sin}^{2} x \: - 1 \: + {sin}^{2}x }$

$= \dfrac{4 {sin}^{3}x \: - \: 2sinx \: }{2 {sin}^{2}x \: - 1 }$

$= \dfrac{2 \: sinx \: ( \: 2 {sin}^{2} x \: - \: 1)}{( \: 2 {sin}^{2}x \: - \: 1 \: )}$

$( \: 2 {sin}^{2} x \: - \: 1 \: ) \: is \: cancel \: out \: from \: numerator \: and \: denominator$

$= 2 \: sinx$

= RHS

Answer 2

${\boxed{\mathtt{\green{To \: Prove}}}}$

$\frac{ \sin(x) - \sin(3x) }{ { \sin(x) }^{2} - { \cos(x) }^{2} } = 2 \sin(x)$

${\boxed{\mathtt{\green{Proof}}}}$

$\mathtt{LHS}$

$= \frac{ \sin(x) - \sin(3x) }{ { \sin(x) }^{2} - { \cos(x) }^{2} }$

➾ Some identities that we'll use in its solution are :-

$> \: \sin(c) - \sin(d) = 2 \cos( \frac{c + d}{2} ). \sin( \frac{c - d}{2} )$

$> \: \cos(2x) = 2 { \cos(x) }^{2} - 1$

$> \: \cos(2x) = 1 - 2 { \sin(x) }^{2}$

➾ Now using these in our LHS .

$= \: \frac{2 \cos( \frac{x + 3x}{2} ) . \sin( \frac{x - 3x}{2} ) }{ \frac{1 - \cos(2x) }{2} - \frac{1 + \cos(2x) }{2} }$

$= \: \frac{2 \cos(2x) . \sin( - x) }{ \frac{1 - \cos(2x) - 1 - \cos(2x) }{2} }$

$= \: \frac{ - 2 \cos(2x) . \sin(x) }{ \frac{ - 2 \cos(2x) }{2} }$

$= \: \frac{2 \cos(2x). \sin(x) }{ \cos(2x) }$

$\boxed{= \: 2 \: \sin(x) }$

$\mathtt{\: LHS\: =\: RHS}$

Hence proved