Proving (tan φ + sec φ - 1)/(tan φ - sec φ + 1) = (1 + sin φ)/cos φ

[chwala]

Homework Statement

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

Homework Equations

The Attempt at a Solution

$(sin φ+1-cos φ)/(sin φ+cos φ-1)$[/B]

 

[mfb]

$(sin φ+1-cos φ)/(sin φ+cos φ-1)$

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]

To deal with expressions $a-1$ in a denominator, it is often useful to expand the quotient by $a+1\,.$

 

[chwala]

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]

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]

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!

 

[chwala]

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 $cos ψ?$

 

[chwala]

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]

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...

 

[chwala]

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

 

[LCKurtz]

Your equations would look much nicer if you used \sin and \cos in your tex expressions.
For example, $\sin\psi$ versus $sin\psi$.

 

[chwala]

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

 

[fresh_42]

I have already mentioned one: expand the quotients by $\cos \varphi + \sin \varphi +1$.