pythegorian identity: tan^2t+1= sec^2

[bin_sina]

please help! can't figure out this- "define pythagorean identity in terms of sec(t) and tan(t)"

is it tan^2t+1= sec^2

and square- rooted both side ..............
finally got something like

tan(t) = square root of sec^2(t)-1
is this the answer? show me how please!

sorry coz i don't don't know how to use math xters

 

[Denis]

tan^2(t) = sec^2(t) - 1
tan(t) = sqrt[sec^2(t) - 1] : brackets are IMPORTANT

 

[bin_sina]

but how are they related?

 

[tkhunny]

It is not real clear what is wanted.

What is the "Pythagorean Identity"? Write it down. Draw yourself a right triangle, mark an angle or two, and plot it out in your mind.

We have a^{2} + b^{2} = c^{2}

Pick an angle, maybe \(\displaystyle \alpha\) between a and c.

So, \(\displaystyle \sin(\alpha)=b/c\) and \(\displaystyle \cos(\alpha)=a/c\).

We also know \(\displaystyle \sin^{2}(\alpha)+\cos^{2}(\alpha)=1\)

Where does that leave us?

 

[bin_sina]

then tan(alpha) should also equal b/a

sec( alpha) =1/sine

in terms of sec,
?

 

[galactus]

sec(x)=1/cos(x), not 1/sin(x)

 

[soroban]

Hello, bin_sina!

I have no idea what the question is asking for . . .

Define Pythagorean identity in terms of \(\displaystyle \sec(t)\) and \(\displaystyle \tan(t)\)

Are they asking us to derive a Pythagorrean identity in terms of \(\displaystyle \sec(t)\) and \(\displaystyle \tan(t)\) ?


\(\displaystyle \text{Start with: }\L\:\sin^2(t)\,+\,\cos^2(t)\;=\;1\)

\(\displaystyle \text{Divide by }\cos^2(t):\;\;\L\frac{\sin^2{t}}{\cos^2(t)}\,+\,\frac{\cos^2(t)}{\cos^2(t)} \;=\;\frac{1}{\cos^2(t)}\)

. . \(\displaystyle \text{then we have: }\L\:\left(\frac{\sin(t)}{\cos(t)}\right)^2\,+\,1\;=\;\left(\frac{1}{\cos(t)}\right)^2\)


\(\displaystyle \text{Therefore: }\L\:\fbox{\tan^2(t)\,+\,1\;=\;\sec^2(t)}\)