Question

Could someone please help me verify the trig. equation using identities?
1+sec^2xsin^2x=sec^2x

Answer 1

you could start by the first part of the equality
\[1+\sec^{2}x*sen^{2}x\]
you have to remember these identities :
\[secx=1/cosx\]
\[sen^{2}x+cos^{2}x=1\]
replacing
\[1+\frac{ 1^{2} }{\cos^{2}x }*sen^{2}x\]
\[\frac{ cos^{2}x+sen^{2}x }{\cos^{2}x}\]
\[\frac{ 1 }{\cos^{2}x}=sec^{2}x\]

Answer 2

Thanks so much! @baad1994 Could you help me with one more?
-5tan^2x+sec^2x=1

Answer 3

And haha, I actually understood what steps you took. (: So thank you so much!!

Answer 4

nice! :) , are you sure that there is a 5 before tan^2x?

Answer 5

Oh, sorry. I meant -tan^2x not 5tan^2x

Answer 6

oh ok , no problem n.n
\[-tg^{2}x+\sec^{2}x=1\]
the same, start by the first part
\[-tg^{2}x+\sec^{2}x\]
Remember these:
1)\[tgx=\frac{ senx }{ cosx }\]
2)\[secx=\frac{ 1 }{ cosx }\]
3)\[sen ^{2}x+\cos ^{2}x = 1 \]
\[\cos ^{2}x = 1-sen ^{2}x \]
replacing
\[-\frac{ senx^{2} }{ cosx^{2} }+\frac{ 1 }{ cosx^{2} }\]
\[\frac{ 1-sen^{2}x }{ \cos ^{2}x }\]

Answer 7

when you are going to verify trigonometric expressions you usually have to pass all to sen and cos. And remember that sen^2x+cos^2x=1

Answer 8

so remember the equivalences:
secx=1/cosx
cscx=1/senx
tgx=senx/cosx
ctgx=cosx/senx
and operate

Answer 9

Thanks so much!! (: