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anonymous
 one year ago
Prove  tan^2(x) + sec^2(x) = 1 by working on one side to match the other using identities.
anonymous
 one year ago
Prove  tan^2(x) + sec^2(x) = 1 by working on one side to match the other using identities.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \tan^2x + \sec^2x = 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \tan^2x + \sec^2x = 1\]

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0sec^2 theta = what ? remember the identity ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0well that's reciprocal of sec but it's okay we can use that too!! tan^2 =what ?

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0yes right so replace tan and sec with that \[\huge\rm \frac{ \sin^2x }{ \cos^2x } +\frac{ 1 }{ \cos^2x}\] find the common denominator

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0ohh well not gonna work should use the identity i guess

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1You could also use the fact that \(\sf sec^2(\theta) = tan^2(\theta) +1\) and then substitute this in place of \(\sf \sec^2(\theta)\)

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[\sf tan^2(\theta)+\sec^2(\theta) = 1\]\[\sf \tan^2(\theta) +\color{red}{\tan^2(\theta) +1}=1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wow... The one identity I didn't think of solved it so easily. Thank you!

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge\rm \frac{ \sin^2x +1}{ \cos^2 }\] use the special identity sin^2x+cos^2x =1 solve for cos^2

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.0here you can copy these identities http://www.math.com/tables/trig/identities.htm you weren't familiar with this so that's why i thought better to write interms of sin and cos

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1That's a good way to approach it too. @Nnesha :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks that will help too @Nnesha
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