Question

1+secA/sec A = sin^2 A / 1-cos A​

Answer 1

Answer: see proof below

Explanation:

`(1+\sec A)/(\sec A)=(\sin^2 A)/(1-\cos A)`

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

`\text{LHS:}\qquad \qquad (1+\sec A)/(\sec A)`

`\text{Identity:}\qquad \qquad (1+(1)/(\cos A))/((1)/(\cos A))`

`\text{Simplify:}\qquad \qquad ((\cos A+1)/(\cos A))/((1)/(\cos A))\\\\\\.\qquad \qquad \qquad =(1+\cos A)/(1)`

`\text{Multiply:}\qquad \qquad (1+\cos A)/(1)\cdot \bigg((1-\cos A)/(1-\cos A)\bigg)\\\\\\.\qquad \qquad \qquad =(1-\cos^2 A)/(1-\cos A)`

`\text{Identity:}\qquad \qquad (\sin^2 A)/(1-\cos A)`

`\text{LHS = RHS:}\quad (\sin^2 A)/(1-\cos A)=(\sin^2 A)/(1-\cos A)\quad \checkmark`