Proving identities !!!

_____1_______ = secO - tanO
secO + tanO

**O means theata

What I did was turn the denominator on the right side into (1/cos) + (sin/cos) and then added them so it became (1+sin/cos).

that was my denominator for the right side ---> 1 / ((1+sin) / cos) = sec - tan

then I multiplied by the reciprocal on the right side of the equal sign because it's a complex fraction

1/1 multiplied by cos / (1+sin)

this becomes cos / (1+sin)

I then multiplied that by cos/cos because it's the same thing as 1 and it will give you a Pythagorean identity

it becomes: cos^2 / (cos)(1+sin)

I turned cos^2 into (1-sin^2) which is a pythag identity
then I factored (1-sin^2) into (1+sin)(1-sin)

that becomes:
(1+sin)(1-sin) / (cos)(1+sin)

the (1+sin) cancels out

(1-sin) / cos

then you just split it into 2 fractions (they both have the same denominator)

(1/cos) - (sin/cos)

which brings you to your final answer

sec - tan = sec - tan

hope it helps :)

Multiply both the numerator and the denominator by (sec O - tan O)
Multiplying out with FOIL, you get (sec O - tan O)/(sec^2 O - tan^2 O)
A trigonometric identity states that sec^2 O = tan^2 O + 1
Rearranging to our needs, sec^2 O - tan^2 O=1.
Therefore the the fraction reduces to (sec O - tan O)/1
=sec O - tan O.

I hope this helps.

_____1_______=secO - tanO
secO + tanO

LHS
multipiy and devide with conjugate
_____1_______ *(secO - tanO )
(secO + tanO )*(secO - tanO )
=secO - tanO /((sec0)2-(tan0)2)
=sec0-tan0 {(sec0)2-(tan0)2)=1

1/(sec + tan)
= cos/(1 + sin)
= cos*(1 - sin)/((1 + sin)(1 - sin))
= cos*(1 - sin)/(1 - sin^2)
= cos*(1 - sin)/(cos^2)
= (cos - cos*sin)/cos^2
= (1 - sin)/cos
= sec - tan