Question

Sec A -1/Sec A -1=(sin A/1+cos A)^2

Answer 1

Step-by-step explanation:

To prove--->

( SecA - 1 ) / ( SecA - 1 ) = ( SinA / 1 + CosA )²

Proof----> LHS

= ( SecA - 1 ) / ( SecA + 1 )

We know that,

SecA = 1 / CosA , using it , we get,

= { ( 1 / CosA ) - 1 } / { ( 1 / CosA ) + 1 }

Taking , CosA , LCM , in , numerator and denominator , we get,

= { ( 1 - CosA ) / CosA } / { ( 1 + CosA ) / CosA }

= ( 1 - CosA ) / ( 1 + CosA )

Multiplying by ( 1 + CosA ) in numerator and denominator , we get,

= ( 1 - CosA ) ( 1 + CosA ) / ( 1 + CosA ) ( 1 + CosA )

We know that,

a² - b² = ( a + b ) ( a - b ) , applying it in numerator , we get,

= { ( 1 )² - ( CosA )² } / ( 1 + CosA )²

= ( 1 - Cos²A ) / ( 1 + CosA )²

We know that,

1 - Cos²A = Sin²Α , applying it here we get,

= Sin²A / ( 1 + CosA )²

= ( SinA / 1 + CosA )² = RHS