Help I have a test tomorrow

[errant flame]

Help I have a test tomorrow I don't understand at all this is trig identities

[worldly ore]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with: ```diff
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[runic bramble]

which part?

[errant flame]

Almost everything for left side

I don't get how they got 1- sin theta divided by 1- sin thetha when multiplying

@runic bramble

[runic bramble]

they make use of the identity:
sin²x+cos²x = 1
=> cos²x = 1-sin²x
=> cos²x = (1+sin x)(1-sin x)
(remember the formula a²-b² = (a+b)(a-b))

[errant flame]

I don't get it

Can you start from the beginning of left side and explain step by step

@runic bramble

[errant flame]

I get most of it now but I don't get this part

[runic bramble]

from the RHS
we know that sec²x = 1/cos²x
and the denominator of the 2nd term is also cos x
so to prove from the LHS
we also need to make the denominator contains cos x
so we use the identity:
sin²x+cos²x = 1
=> cos²x = 1-sin²x
=> cos²x = (1+sin x)(1-sin x)
since it's the most suitable identity to make the denominator contains cos x
so after multiplying we get:
(1/(1+sin x))((1-sin x)/(1-sin x))
= (1-sin x)/(1-sin²x)
(use the formula (a+b)(a-b) = a²-b² for the denominator)
= (1-sin x)/cos²x
then split it into 2 fractions:
(1/cos²x)-(sin x/cos²x)
= sec²x-(sin x/cos x)(1/cos x)
we know that tan x = sin x/cos x, so:
= sec²x-(tan x)(1/cos x)
= sec²x-(tan x/cos x)