Question

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Answer 1

`\text{L.H.S}\\\\=(1)/(\sin^2 A) - \frac1{\tan^2 A}\\\\\\=(1)/(\sin^2 A)- \frac{1}{\tfrac{\sin^2 A}{\cos^2 A}}\\\\\\=(1)/(\sin^2 A) - (\cos^2 A)/(\sin^2 A)\\\\\\=(1 - \cos^2 A)/(\sin^2 A)\\\\\\=(\sin^2 A)/(\sin^2 A)\\\\\\=1\\\\=\text{R.H.S}\\\\\text{Proved.}`