Question

1+sin theta divided by 1-sin theta= (sec theta+tan theta) whole square

Answer 1

Theta is written as A.

Answer:

( 1 + sinA ) / ( 1 - sinA ) = ( secA + tanA )^2

Step-by-step explanation:

Left Hand Side :

= > ( 1 + sinA ) / ( 1 - sinA )

Dividing and multiply by ( 1 + sinA ) :

= > { ( 1 + sinA ) / ( 1 - sinA ) } x { ( 1 + sinA ) / ( 1 + sinA ) }

= > { ( 1 + sinA )^2 } / { ( 1 - sinA ) ( 1 + sinA ) }

= > { ( 1 + sinA )^2 } / { ( 1 )^2 - ( sinA )^2 } { Using ( a + b )( a - b ) }

= > { ( 1 + sinA )^2 } ) { 1 - sin^2 A }

From the properties of trigonometry :

• 1 - sin^2 B = cos^2' B

= > { ( 1 + sinA )^2 } / ( cos^2 A )

= > [ ( 1 + sinA ) / cosA ]^2

= > [ 1 / cosA + sinA / cosA ]^2

= > ( secA + tanA )^2

Left Hand Side = Right Hand Side.

Hence, proved.