Question

how do i prove this????? pls help
- tan2x + sec2x = 1

Answer 1

-tan2x+sec2x
\[\frac{ -\sin2x }{ \cos2x }+\frac{ 1 }{ \cos2x }\]\[\frac{ -\sin2x+1 }{ \cos2x }\]\[1-\sin2x=\cos2x\]so\[\frac{ \cos2x }{ \cos2x }=1\]..proven...

Answer 2

thank you sooo much @LynFran

Answer 3

ur welcome..please medal..thanks

Answer 4

@DecentNabeel its the Pythagorean identity\[cosx+sinx=1\]so \[\cos2x+\sin2x=1 \]
so \[\cos2x=1-\sin2x\]

Answer 5

@freckles please tell @DecentNabeel im right

Answer 6

I'm gonna have to agree with @DecentNabeel here
but I do think he meant to write
\[-\tan^2(x)+\sec^2(x)=1 \text{ instead of } -\tan(2x)+\sec(2x)=1\]

Answer 7

that last I was thinking about the OP when I said he

Answer 8

\[-\tan ^2x+\sec ^2x=1\]

Answer 9

Is that^^^what you are trying to prove?

Answer 10

\[\sin(x)+\cos(x) \neq 1 \text{ for all } x \\ \text{ the pythagorean identity is } \\ \sin^2(x)+\cos^2(x)=1\]

we see here the following:
\[\cos(\theta)=\frac{x}{r} \implies r \cos(\theta)= x \\ \sin(\theta)=\frac{y}{r} \implies r \sin(\theta)= y \\ \text{ now by Pythagorean theorem we have } \\ x^2+y^2=r^2 \\(r \cos(\theta))^2+(r \sin(\theta))^2=r^2 \\ r^2 \cos^2(\theta)+r^2 \sin^2(\theta)=r^2 \\ \text{ divide both sides by } r^2 \\ \cos^2(\theta)+\sin^2(\theta)=1 \]