Question

how do you solve sec^2x=tanx+1

Answer 1

1st step...replace sec^2(x) with tan^2(x) + 1

Answer 2

Okkk so here is how I wld solve this:

we know that \( \sec^2x= \large \frac{1}{\cos^2x}\)

Answer 3

and \( \tan x = \large \frac{ \sin x}{ \cos x}\)

so \[ \sec^2x= \tan x +1 \]
\[\frac{1}{ \cos ^2x}=\frac{ \sin x}{ \cos x}+1 \]
Now we will multiply both sides of the equation by \( \cos ^2 x \) and we get:
\[1= \sin x \cos x +\cos^2x\]
One of the trig identities is \( 1= \sin^2x+ \cos^2x\)

Therefore we replace the 1 with the identity and get:
\[\sin^2x +\cos^2x=\sin x \cos x +\cos^2x\]

We subtract \( \cos^2 x\) from both sides and get:
\[\sin^2x=\sin x +\cos x \]
Then we divide both sides by \( \sin x\) and
\[ \sin x=\cos x\]
Then we divide both sides by \( \cos x \) and finally get
\[ \frac{ \sin x}{ \cos x} =1\]
Which is \( \tan x =1 \)

Then we multiply by sides by inverse tan and get:
\[ \tan ^{-1} x=1\]

Ummm and as you know at \( \tan (45)=1\)
So our x=45 degrees

Answer 4

Do you follow?

Answer 5

thanks BlackLabel

Answer 6

Ummm do you follow what i did?
Any questions?