Question

How do i prove this identity: sin^2x + cos^2x + tan^2x = 1/cos^2x?

Answer 1

sin^2x + cos^2x = 1

1 + tan^2 x = 1/cos^2 x

multiply everything by cos^2 x

cos^2 x + cos^2x (sin^2 x/cos^2 x) = 1

cos^2 x + sin^2 x = 1

which is the property we used a while ago

Answer 2

\[\sin ^{2}x + \cos ^{2}x + \tan ^{2}x = 1/\cos ^{2}x\]

In the end I got \[1 + 1-\cos ^{2}x \div \cos ^{2}x = 1divcos ^{2}\]

Answer 3

\[\Large \sin^2x + \cos^2x + \frac{\sin^2x }{\cos^2x}\]Now it looks fine.

Answer 4

although i dont know if that was valid since i looped

Answer 5

how? What are your steps? I'm so confused D;

Answer 6

we can use sef's step too

Answer 7

alright so from sin^2x + cos^2x + tan^2x = 1/cos^2x let us simplify everything so first lets start from the left side sin^2x + cos^2x + tan^2x = 1/cos^2x?

we know sin^2x + cos^2x = 1 and tan^2x =sin^2x/cos^2x

so therefore we get 1 + sin^2x/cos^2x and now if we find the common denominator we get (cos^2x +sin^2x)/cos^2x = 1/cos^2x which is exactly the same as the right side :)

Answer 8

\[\Large \sin^2 x + \cos^2x =1\]
\[\Large 1+\frac{sin^2x}{cos^2x}\]
\[\Large \frac{1(cos^2x) +sin^2x}{cos^2x}\]
\[\Large \frac{\cos^2x + \sin^2x }{\cos^2x}\]
Again,
\[\Huge \frac{1}{\cos^2x}\]

Answer 9

so basically the trick to theses questions are if you see tan convert to sin or cos and if its fractions do common denominator and what not :) hope you understand

Answer 10

\[\sin^2 x + \cos^2 x + \frac{\sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x}\]

\[\sin^2 x + \cos^2 = \frac{1}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x}\]

\[\sin^2 x + \cos^2 x = \frac{1 - \sin^2 x}{\cos^2 x}\]

\[\sin^2 x + \cos^2 x = \frac{\cos^2 x}{\cos^2 x}\]

\[\sin^2 x + \cos^2 x = 1\]

this is a property

Answer 11

ohhh... I think I'm getting it, thanks guys!