Question

Prove that :1/sin^2A -1/tan^2A =1​

Answer 1

To prove that:

$\frac{1}{\sin ^{2}A}-\frac{1}{\tan ^{2}A}=1$

LHS = $\frac{1}{\sin ^{2}A}-\frac{1}{\tan ^{2}A}$

Using basic trigonometric identity, $\tan (x)=\frac{\sin (x)}{\cos (x)}$

       $=\frac{1}{\sin ^{2}A}-\frac{1}{\left(\frac{\sin A}{\cos A}\right)^{2}}$

      $=\frac{1}{\sin ^{2}A}-\frac{1}{\frac{\sin^2A}{\cos^2 A}}$

      $=\frac{1}{\sin ^{2}A}-\frac{\cos^2 A}{\sin^2A}$

      $=\frac{1-{\cos^2 A}}{\sin ^{2}A}$

Using trigonometric identity: $1-\cos ^{2}(x)=\sin ^{2}(x)$

      $=\frac{\sin ^{2}A}{\sin ^{2}A}$

      = 1

      = RHS

LHS = RHS

$\frac{1}{\sin ^{2}A}-\frac{1}{\tan ^{2}A}=1$

Hence proved.