Question

Prove that √(1+sin A/1-sin A) =sec A +tana

Answer 1

Step-by-step explanation:

given that ; √(1+sin A /1-sin A) = sec A + tan A L.H.S = [√( 1+sin A / 1- sin A) ] rationalize with [√(1+sin A/ 1+sin A) ] = [√(1+sin A)/(1-sin A) ] [√1+sin A/1+sin A)] = [√[( 1+sin A)^2/ 1-sin^2A ] ] = (1+sin A / cos A) = ( 1/cos A ) + (sin A / cos A ) = sec A + tan A . there fore L.H.S = R.H.S

Answer 2

Step-by-step explanation:

first take the LHS=√(1+SinA)/(1-SinA)

rationalizing the denominator=√(1+Sin A)/(1-Sin A)*√(1+Sin A)/√(1+ Sin A)

=√(1+SinA)^2/√(1-Sin^2 A)  [square and root will get cancelled in                                 numerator] [ in denominator identity will be used sin^2 A +Cos^2A=1]

Therefore, = 1+Sin A/Cos A.   [Square of cos^2 A cancelled as it was under root]

=1/Cos A +Sin A/ Cos A

=Sec A +Tan A

Hence proved. Hope you will find it helpful.