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  • 10-04-2022
  • Mathematics
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loneguy
loneguy loneguy
  • 10-04-2022

[tex]\text{L.H.S}\\\\=\dfrac{1}{\sin^2 A} - \dfrac1{\tan^2 A}\\\\\\=\dfrac{1}{\sin^2 A}- \dfrac{1}{\tfrac{\sin^2 A}{\cos^2 A}}\\\\\\=\dfrac{1}{\sin^2 A} - \dfrac{\cos^2 A}{\sin^2 A}\\\\\\=\dfrac{1 - \cos^2 A}{\sin^2 A}\\\\\\=\dfrac{\sin^2 A}{\sin^2 A}\\\\\\=1\\\\=\text{R.H.S}\\\\\text{Proved.}[/tex]

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