# Trigonometry identities

Gold Member

## Homework Statement

Show that
## {(tan φ+sec φ-1)/(tan φ-sec φ+1)}≡ {(1+sin φ)/cos φ}##[/B]

## The Attempt at a Solution

## (sin φ+1-cos φ)/(sin φ+cos φ-1)##[/B]

mfb
Mentor
## (sin φ+1-cos φ)/(sin φ+cos φ-1)##
How did you get this?
It is impossible to tell where you need help if you don't show the steps you took so far.

fresh_42
Mentor
2021 Award
To deal with expressions ##a-1## in a denominator, it is often useful to expand the quotient by ##a+1\,.##

Gold Member
To deal with expressions ##a-1## in a denominator, it is often useful to expand the quotient by ##a+1\,.##
That is what i did....let me look at my working again...

fresh_42
Mentor
2021 Award
No, you expanded the entire thing by ##\cos \varphi##. Now you can go ahead and write the denominator as ##a-1## and make ##a^2-1## out of it.

chwala
Gold Member
No, you expanded the entire thing by ##\cos \varphi##. Now you can go ahead and write the denominator as ##a-1## and make ##a^2-1## out of it.
still not getting...how can you get ##a-1## in denominator?
look at my work now
##{(tan ψ+sec ψ-1)(tan ψ-sec ψ-1)}/{(tan ψ-sec ψ+1)(tan ψ-secψ-1)}##
let ## b= sec ψ-1##
we have
##{(tan ψ+b)(tanψ+b)}/{(tan^2ψ-b^2ψ)}##
after cancelling## (tan ψ+b)##
we get the original problem again!

Gold Member
No, you expanded the entire thing by ##\cos \varphi##. Now you can go ahead and write the denominator as ##a-1## and make ##a^2-1## out of it.
not really...how ## cos ψ?##

Gold Member
anyway, let me try do it and post my solution, i believe i am capable...then you can post alternative way of doing it..

symbolipoint
Homework Helper
Gold Member
The lefthand member uses tangents and secants. The righthand member uses sines and cosines. Try starting with definition of tangent and secant on the left side; and see what other simplifications and algebraic steps you can pick...

Gold Member
I nailed it, i guess sometimes i am just too tired or not motivated. Here
## {(sin ψ+1-cosψ)/cosψ}.{(cos φ)/sin φ-1+cosφ)}##
##{(sin φ+1-cosφ)cos φ)/(sin φ-1+cosφ)cos φ)}##
##{sin ψcosψ+cosψ-cos^2ψ)/(sin φ-1+cosφ)cos φ)}##
##{sin ψcosψ+cosψ-1+sin^2ψ)/(sin φ-1+cosφ)cos φ)}##
##{(sin^2ψ-1+cosψ+sinψcosψ)/(sin φ-1+cosφ)cos φ)}##
##{(sin φ+1)(sinφ-1)+cosφ(1+sinφ)/(sin φ-1+cosφ)cos φ)}##
##{[(sin φ+1)][(sinφ-1+cosφ)]/(sin φ-1+cosφ)cos φ)}##
##(sin φ+1)/cosφ##

Last edited:
LCKurtz
Homework Helper
Gold Member
Your equations would look much nicer if you used \sin and \cos in your tex expressions.
For example, ##\sin\psi## versus ##sin\psi##.

Gold Member
There are 2 other methods to this....from my colleagues, i can share...

fresh_42
Mentor
2021 Award
I have already mentioned one: expand the quotients by ##\cos \varphi + \sin \varphi +1##.

chwala