verifying identities

[jmccomas]

sin sq. x +tan sq. x +co sq. x = sec sq. x
I having trouble finding a starting point.

 

[galactus]

\(\displaystyle sin^{2}(x)+tan^{2}(x)+cos^{2}(x)=sec^{2}(x)\).

Just replace \(\displaystyle tan^{2}(x)\) with \(\displaystyle sec^{2}(x)-1\) and replace \(\displaystyle cos^{2}(x)\) with \(\displaystyle 1-sin^{2}(x)\), then do the arithmetic.

It will pop out at you.

 

[soroban]

Hello, jmccomas!

A slightly different explanation . . .

\(\displaystyle \text{Prove: }\:\sin^2\!x + \tan^2\!x + \cos^2\!x \;=\; \sec^2\!x\)

\(\displaystyle \text{We have: }\:\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} + \tan^2\!x \;\;=\;\;\underbrace{1 + \tan^2\!x}_{\text{This is }\sec^2\!x} \;\;=\;\;\sec^2\!x\)

 

[galactus]

That's better yet, Soroban. Duh, I completely overlooked the most obvious, and most famous, identity.

 

[jmccomas]

How about this?
sin^2 x + tan^2 x + cos^2 x = sec^2 x
=(1 - cos^2 x) + cos^2 x + tan^2 X = 1 + tan^2 x
=1 + tan^2 x = sec^2 x

 

[lookagain]

How about this?
sin^2 x + tan^2 x + cos^2 x = sec^2 x
=(1 - cos^2 x) + cos^2 x + tan^2 X = 1 + tan^2 x
=1 + tan^2 x = sec^2 x

jmccomas,

no, keep the expression on one (the same) side throughout your steps.
Also, in your second line, it's as if you have done scratch work
and inserted that as a step. If so, that is not appropriate. In any event,
you would not have your second line as you typed.

Instead, if you intend to go with your route, you could show:


\(\displaystyle \sin^2x + \tan^2x + \cos^2x = \sec^2x\)

\(\displaystyle \sin^2x + \cos^2x + \tan^2x = \sec^2x\)

\(\displaystyle (1 - \cos^2x) + \cos^2x + \tan^2x = \sec^2x\)

\(\displaystyle 1 - \cos^2x + \cos^2x + \tan^2x = \sec^2x\)

\(\displaystyle 1 + \tan^2x = \sec^2x\)

\(\displaystyle \sec^2x = \sec^2x\)



There is the uninterrupted flow of steps, as each newer
equation can be justified from the equation just prior
to it.

 

[jmccomas]

Understood. Thank you very much.