Question

1/tan^2 x +1

Answer 1

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

Answer 2

\[\frac{ 1 }{ \tan ^{2}x-1 }\]

Answer 3

how do you get 1/sec^2x ?

Answer 4

do you know the identity sin^2x + cos^x = 1

Answer 5

yes

Answer 6

1 + tan^2 x
1 + sin^x/cos^x
(cos^2x + sin^2x)/cos^2x
1/cos^2x
sec^2 x

hence 1 + tan^2 x = sec^2 x

Answer 7

but dont you have to multiply (cos^2 x+sin^2 x) by cos^2 x to get the same numerator on both sides so you could place it all in one thing?

Answer 8

I didn't get what you want to ask

Answer 9

like this is where im getting confused
\[\frac{ 1 }{ \tan ^{2}x-1}\]
\[\frac{ 1 }{ \frac{ \sin ^{2}x }{ \cos ^{2}x } +\sin^{2}x +\cos ^{2}x }\]
\[\frac{ 1 }{ \frac{ \sin ^{2}x }{ \cos ^{2}x }+\frac{ \cos ^{2}x (\sin ^{2}x +\cos ^{2}x) }{\cos ^{2}x } }\]

Answer 10

this is as far as i get

Answer 11

ho w you write second step that is wrong

Answer 12

both sign will be minus
- sin^2 x - cos ^ 2x

Answer 13

well what if i didny i wrote it wrong it is a + in bettween tan^2 and 1
making
1/tan^2 + 1

Answer 14

ok then you will get
in numerator cos^2 x
in denominator sin^2 x + cos^2 x *1
that will be sin^2 x + cos^2 x
hence denominator will be 1
so you get cos^2x
that can be written as 1/ sec^2 x

Answer 15

Start with Pythagorean Identity
\[1 = \sin^2x + \cos^2x \]

Divide each side by \(\cos^2x\):

\(\dfrac{1}{\cos^2x} = \dfrac{\sin^2x}{\cos^2x} + \dfrac{\cos^2x}{\cos^2x}\)

Recall the rule \(\dfrac{a^2}{b^2} = \left(\dfrac{a}{b}\right)^2\)

So, it follows that:

\(\left(\dfrac{1}{\cos x}\right)^2 = \left(\dfrac{\sin x}{\cos x}\right)^2 + \left(\dfrac{\cos x}{\cos x}\right)^2\)

So therefore that simplifies to

\(\sec^2x = \tan^2x + 1\)

Answer 16

that makes sense to thank you hero :)

Answer 17

and thank you rishi as well for everything and all your help