Question

Prove :- 1+sinA \1- sinA = (sec A +tan A )^2​

Answer 1

Answer:

[ ( 1 + sin A ) / ( 1 - sin A ) ] = ( sec A + tan A )²

Step-by-step-explanation:

The given trigonometric expression is

[ ( 1 + sin A ) / ( 1 - sin A ) ] = ( sec A + tan A )²

We have to prove this expression.

Now,

[ ( 1 + sin A ) / ( 1 - sin A ) ] = ( sec A + tan A )²

∴ RHS = ( sec A + tan A )²

We know that,

( a + b )² = a² + b² + 2ab

⇒ RHS = sec² A + tan² A + 2 sec A tan A

We know that,

sec A = 1 / cos A

⇒ RHS = ( 1 / cos² A ) + tan² A + 2 * ( 1 / cos A ) tan A

We know that,

tan A = sin A / cos A

⇒ RHS = ( 1 / cos² A ) + sin² A / cos² A + 2 * ( 1 / cos A ) * ( sin A / cos A )

⇒ RHS = [ ( 1 + sin ² A ) / cos² A ] + ( 2 sin A / cos² A )

⇒ RHS = [ ( 1 + sin² A + 2 sin A ) / cos² A ]

⇒ RHS = [ ( 1 + sin A )² / cos² A ]

We know that,

sin² A + cos² A = 1

⇒ cos² A = 1 - sin² A

⇒ RHS = [ ( 1 + sin A )² / ( 1 - sin² A ) ]

We know that,

a² - b² = ( a + b ) ( a - b )

⇒ RHS = [ ( 1 + sin A ) ( 1 + sin A ) / ( 1 + sin A ) ( 1 - sin A ) ]

⇒ RHS = [ ( 1 + sin A ) / ( 1 - sin A ) ]

LHS = [ ( 1 + sin A ) / ( 1 - sin A ) ]

∴ LHS = RHS

Hence proved!