Notes: Prove that (1 + sinA) / CosA = (sinA- cosA+1) / (sinA+ cosA -1 )any proof with out cross multiplication

Sunday, September 28, 2014

We'll use properties of proportions to prove the
identity.

First, we'll use componendo
property:

$\displaystyle\frac{{a}}{{b}}=\frac{{c}}{{d}}=\gt\frac{{{a}+{b}}}{{b}}=$
(c+d)/d

$\displaystyle\frac{{{1}+{\sin{{A}}}+{\cos{{A}}}}}{{\cos{{A}}}}=$
(sinA-cosA+1+sinA+cosA-1)/(sinA+cosA-1)

We'll reduce like
terms from the numerator from the right side:

$\displaystyle{\left({1}+{\sin{{A}}}+\right.}$
cosA)/cosA= (2sinA)/(sinA + cosA - 1)

Now, we'll cross
multiply:

(sinA+cosA + 1)(sinA + cosA - 1) =
2sinAcosA

We notice that the product from the left side
returns a difference of two squares, while, to the right side, we've get a double angle
identity:

$\displaystyle{{\left({\sin{{A}}}+{\cos{{A}}}\right)}}^{{2}}-{1}={\sin{}}$
2A

We'll expand the
binomial:

$\displaystyle{{\sin}}^{{2}}{A}+{2}{\sin{{A}}}{\cos{{A}}}+{{\cos}}^{{2}}{A}-{1}={\sin{}}$
2A

But, from Pythagorean identity, we'll get $\displaystyle{{\sin}}^{{2}}{A}+$
cos^2A = 1 $\displaystyle.$

1 + sin2A - 1 =
sin2A

We'll eliminate like terms from the left
side:

sin2A =
sin2A

We notice that we've get the
same double angle identity both sides, therefore, the given expression represents an
identity.

Notes