Question

Please Prove L.H.S=R.H.S.(Give steps in detail)and please write and solve

Answer 1

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Answer 2

HEY buddy.....!! here is ur answer

$We \: have \: to \: prove \: that \: : \tan\alpha - \cot\alpha = \frac{2 { \sin }^{2} \alpha - 1 }{ \sin \alpha \cos\alpha } \\ \\ On \: taking \: R.H.S. : \\ \\ = > \frac{2 { \sin }^{ 2} \alpha - 1}{ \sin \alpha \cos \alpha } = - \frac{1 - 2 { \sin }^{2} \alpha }{ \sin \alpha \cos \alpha} \\ \\ = > - \frac{1 - { \sin }^{2} \alpha - { \sin }^{2} \alpha }{ \sin\alpha \cos\alpha } \\ \\ = > - \frac{ { \cos }^{2} \alpha - { \sin }^{2} \alpha }{ \sin\alpha \cos \alpha } \\ \\ As \: we \: know \: that \: : { \sin }^{2} \alpha + { \cos}^{2} \alpha = 1 = > 1 - { \sin}^{2} \alpha = { \cos}^{2} \alpha \\ \\ = > \frac{ { \sin }^{2} \alpha - { \cos }^{2} \alpha }{ \sin \alpha \cos \alpha } \\ \\ = > \frac{ { \sin}^{2} \alpha }{ \sin \alpha \cos \alpha } - \frac{ { \cos }^{2} \alpha }{ \sin\alpha \cos\alpha} \\ \\ = > \frac{ \sin \alpha }{ \cos \alpha } - \frac{ \cos \alpha }{ \sin \alpha } \\ \\ = > \tan \alpha - \cot\alpha = L.H.S. HENCE PROVED... \\ \\ As \: we \: know \: that \: \frac{ \sin \alpha }{ \cos \alpha } = \tan\alpha \: and \: \frac{ \cos\alpha}{ \sin \alpha } = \cot\alpha$

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