Question

Sin2x(Cot^2x-tan^2x)=4cos2x​

Answer 1

Correct Question To Prove :—

$\boxed{ \rm \sin2x( { \cot }^{2}x - { \tan }^{2}x) = 4 \cot2x }$

Proof :—

$\tt \red{L.H.S.} \\ \\ \sf = \sin2x( { \cot}^{2} x - { \tan }^{2} x) \\ \\ = \sf \sin2x \left[ \frac{ { \cos }^{2} x}{ { \sin}^{2} x} - \frac{ { \sin }^{2}x }{ { \cos }^{2}x } \right] \\ \\ \sf = \sin2x \left[ \frac{ { \cos }^{4}x - { \sin}^{4} x }{ { \sin}^{2} x \: { \cos}^{2}x } \right] \\ \\ \sf = 2( \sin x \: \cos x) \left[ \frac{ {( { \cos}^{2} x)}^{2} - {( { \sin}^{2} x )}^{2} }{ {( \sin x \: \cos x )}^{2} } \right] \\ \\ \sf = 2 \cancel{( \sin x \: \cos x)} \left[ \frac{( { \cos }^{2}x - { \sin }^{2}x)( { \cos }^{2} x+ { \sin}^{2} x) }{ \cancel{( \sin x \: \cos x)}( \sin x \: \cos x)} \right] \\ \\ \sf = 2 \left[ \frac{( { \cos}^{2}x - { \sin }^{2} x )(1)}{( \sin x\: \cos x)} \right] \\ \\ \sf = 2 \times 2 \times \frac{1}{2} \times \left[ \frac{ \cos2x}{ \sin x \: \cos x } \right] \\ \\ \sf = 4 \left[ \frac{ \cos2x }{2 \sin x \: \cos x } \right] \\ \\ \sf = 4 \left\{ \frac{ \cos2x}{ \sin2x} \right\} \\ \\ \sf = 4( \cot x) \\ \\ \bf = 4 \cot x \quad\tt = \green{R.H.S.}$

Identities Used :—
$\boxed{ \begin{minipage}{4.5cm} \\ \implies \sin2x = 2 \sin x \: \cos x \\ \\ \implies a^2 - b^2 = (a - b)(a + b) \\ \\ \implies \cos2x = \cos ^2 x - \sin^2 x \\ \\ \implies \cos^2 x + \sin^2 x = 1 \\ \end{minipage} }$

$\mid \underline{\underline{\LARGE\bf\green{Brainliest \: Answer}}}\mid$

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