what have u taken as hypotenuse of triangle is it 1 unit
This is probably a basic arithmetic problem but I have the following to validate
1/(sec + tan) = sec - tan
I simplify to
cos/(sin + 1) = (1-sin)/cos
clearing fractions gives some pythagorean statements which are true, but I bet there's some property about fractions that should tell me the above is true, ie
a/(b+1) = (1-b)/a
How does one verify/validate the identity?
Actually, a proper proof of an identity manipulates only one side of the identity to be proved; otherwise you are assuming what is to be proved. Now you can do what you did as a method for finding a proof, but it is invalid in the proof itself. Furthermore, your proposed property about fractions is generally false.
$a = 0.4\ and\ b = 0.7 \implies \dfrac{0.4}{0.7 + 1} = \dfrac{4}{17}\ and\ \dfrac{0.7 - 1}{0.4} = -\ \dfrac{3}{4}.$ Fractions have different signs are certainly not equal.
The proposed property is true only if $a \ne 0,\ b \ne - 1,\ and\ a^2 = - (b^2 - 1) \implies a^2 = - (b + 1)(b - 1) \implies$
$a = \dfrac{-(b + 1)(b - 1)}{a} \implies \dfrac{a}{b + 1} = \dfrac{- (b - 1)}{a} \implies \dfrac{a}{b + 1} = \dfrac{1 - b}{a}.$
Now getting to the identity.
$\dfrac{1}{sec(x) + tan(x)} = \dfrac{1}{\dfrac{1}{cos(x)} + \dfrac{sin(x)}{cos(x)}} = \dfrac{1}{\dfrac{1 + sin(x)}{cos(x)}} = \dfrac{cos(x)}{1 + sin(x)}.$ This is correct, but may not be helpful.
$\dfrac{1}{sec(x) + tan(x)} = \dfrac{1}{sec(x) + tan(x)} * 1 = \dfrac{1}{sec(x) + tan(x)} * \dfrac{sec(x) - tan(x)}{sec(x) - tan(x)} = \dfrac{sec(x) - tan(x)}{sec^2(x) - tan^2(x)} =$
$\dfrac{sec(x) - tan(x)}{\left(\dfrac{1}{cos(x)}\right)^2 - \left(\dfrac{sin(x)}{cos(x)}\right)^2} = \dfrac{sec(x) - tan(x)}{\dfrac{1}{cos^2(x)} - \dfrac{sin^2(x)}{cos^2(x)}} = \dfrac{sec(x) - tan(x)}{\dfrac{1 - sin^2(x)}{cos^2(x)}} = what?$
I'm afraid trig identities usually revolve around trig functions, not simple arithmetic properties except for multiplying by 1 and adding 0. They take a bit of creativity.
Yes. The trick of multiplying by a strange form of 1 works in this case, but it won't work for every identity. How did I find the right trick in this case.
I cross-multiplied from what was to be proved $\dfrac{1}{sec(x) + tan(x)} = sec(x) - tan(x)$ and found $sec^2(x) - tan^2(x) = 1,$ which is true.
Now I CAN'T actually cross multiply like that in the proof because that's assuming what needs to be proved.
But I can get the same effect on the denominator by multiplying by $\dfrac{sec(x) - tan(x)}{sec(x) - tan(x)} = 1.$
This kind of backwards thinking can be traced back to Proclus, and it probably goes way back farther.