To prove that:
[tex]$\frac{1}{\sin ^{2}A}-\frac{1}{\tan ^{2}A}=1[/tex]
LHS = [tex]\frac{1}{\sin ^{2}A}-\frac{1}{\tan ^{2}A}[/tex]
Using basic trigonometric identity, [tex]\tan (x)=\frac{\sin (x)}{\cos (x)}[/tex]
[tex]$=\frac{1}{\sin ^{2}A}-\frac{1}{\left(\frac{\sin A}{\cos A}\right)^{2}}[/tex]
[tex]$=\frac{1}{\sin ^{2}A}-\frac{1}{\frac{\sin^2A}{\cos^2 A}}[/tex]
[tex]$=\frac{1}{\sin ^{2}A}-\frac{\cos^2 A}{\sin^2A}[/tex]
[tex]$=\frac{1-{\cos^2 A}}{\sin ^{2}A}[/tex]
Using trigonometric identity: [tex]1-\cos ^{2}(x)=\sin ^{2}(x)[/tex]
[tex]$=\frac{\sin ^{2}A}{\sin ^{2}A}[/tex]
= 1
= RHS
LHS = RHS
[tex]$\frac{1}{\sin ^{2}A}-\frac{1}{\tan ^{2}A}=1[/tex]
Hence proved.
