Prove that?

#sqrt((1-sinA)/(1+sinA)) = secA -tanA#
Your first aim should be to get rid of the square root, by making everything within it squared. You'll need to use some identities for this.

#sqrt((1-sinA)/(1+sinA))#

#= sqrt( (1-sinA)/(1+sinA) *(1-sinA)/(1-sinA))#

#= sqrt( (1-sinA)^2/((1+sinA)(1-sinA)))#

Applying #(a+b)(a-b)=a^2 -b^2# for the denominator...

#= sqrt( (1-sinA)^2/(1-sin^2A) )#

Now the numerator is squared, but the denominator still needs to changed using the identity #1-sin^2A = cos^2A#

#= sqrt( (1-sinA)^2/(cos^2A))#

The squares can cancel out the square root...

#= (1-sinA)/(cosA)#

#= 1/cosA -sinA/cosA#

#= secA -tanA#

Here is another way to solve the Problem :

We know that, #1-sin^2A=cos^2A,#

#:. (1-sinA)(1+sinA)=cosA*cosA,#

#:. (1-sinA)/cosA=cosA/(1+sinA)................(star_1).#

But, #(1-sinA)/cosA=1/cosA-sinA/cosA=secA-tanA...(star_2).#

Thus, we conclude from #(star_1) and (star_2),# that,

#secA-tanA=(1-sinA)/cosA, and, secA-tanA=cosA/(1+sinA).#

Multiplying the corresponding sides, we get,

#(secA-tanA)(secA-tanA)=(1-sinA)/cancelcosA*cancelcosA/(1+sinA),#

#i.e., (secA-tanA)^2=(1-sinA)/(1+sinA),# equivalently,

# secA-tanA=sqrt{(1-sinA)/(1+sinA)}.#

Enjoy Maths.!