Trig/ Pre Cal. How do I do this? I need to work out these using Trig identities \( tan^2 + 5 = sec^2 + 4 \) \( \frac{sin^2}{cos^2} + 5 = sec^2 + 4 \) Am I on the right path? What should I do next?

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Trig/ Pre Cal. How do I do this? I need to work out these using Trig identities \( tan^2 + 5 = sec^2 + 4 \) \( \frac{sin^2}{cos^2} + 5 = sec^2 + 4 \) Am I on the right path? What should I do next?

Mathematics
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well....i think we can write 5 as 1+4
\[\huge\rm tan^2 +5 = \tan^2 +1+4\] like this then use this identity 1+tan^2 = sec^2
i guess.....

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\( \frac{1-cos^2}{1-sin^2} + 5 = sec^2 + 4 \) Is this the right path? How do I get the left side to = \( sec^2 + 4 \)
if you do the other way like i did tan^2 +1+4 now you can apply this identity (1+tan^2) =sec^2 = done! 2 steps
tan^2+1+4 = tan^2+5 but I need tan^2 + 5 to equal sec^2+4 How does tan^2 become sec^2? Only 1+tan^2 = sec^2
yes that's right that's why i wrote 1+4 instead 5 now you can apply that identity\[\huge\rm \color{reD}{tan^2 +1}+4 = sec^2 +4 \]
tan^2 + 1 is same as 1+tan^2 \[\huge\rm \color{reD}{1+tan^2 }+4 = sec^2 +4 \]
I have tan^2 on the left said. You are saying tan^2 = 1+tan^2. You just through out tan^2 and put in 1+ tan^2 but tan^2 does not = 1+tan^2
nope i never said that tan^2 = 1+tan^2 well start it again \[\huge\rm tan^2 +5\] is same as \[\tan^2 +1+4\] agree or no ??
1+4 =5 you can write 5 as 1+4
AH I see!!! Man that was right in front of me.. I was way over thinking this thing lol
yaya!!! :-)
i can understand i guess helping other users and PRACTICE help me to find out easy way :-)
just practice practice and practice :-)
Thank you for showing me the light :-)
haha my pleasure!!! :-)

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