Question

prove 1+secA/secA=sin^2A/1-cosA

Answer 1

Hi ,

LHS = ( 1 + secA )/secA

= ( 1 + 1/cosA ) / ( 1/cosA )

= [ ( CosA + 1 ) / cosA ] / ( 1/ cosA )

= Cos A + 1

= ( 1 + cosA ) ( 1 - cosA ) / ( 1 - cosA )

= ( 1 - cos² A ) / ( 1 - cosA )

= Sin² A / ( 1 - cosA )

= RHS

Answer 2

Answer and Explanation:

To prove : $\frac{1+\sec A}{\sec A}=\frac{\sin^2 A}{1-\cos A}$

Proof :

Taking LHS,

$LHS=\frac{1+\sec A}{\sec A}$

Substitute, $\sec A=\frac{1}{\cos A}$

$LHS=\frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$    

$LHS=\frac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}$

$LHS=\frac{\cos A+1}{\cos A}\times \cos A$

$LHS=\cos A+1$

Multiply and divide by $\cos A-1$

$LHS=(\cos A+1)\times \frac{\cos A-1}{\cos A-1}$

$LHS=\frac{(\cos A+1)(\cos A-1)}{\cos A-1}$

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

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

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

$LHS=RHS$

Hence proved.