Question

Verify the following identity

Answer 1

well i have done it but i dont nkow how to put it here .. let me give it a try

Answer 2

Do you use MathType or equation editor. You can do it there and take a screenshot to post... Thanks...

Answer 3

lol well i never used b4 just new here but let me give it a try

Answer 4

multiply by the base on both sides an apply the identities tan(x)=sin(x)/cos(x) and sec(x)=1/cox(x)

Answer 5

@nubber maybe u can use draw

Answer 6

\[{1\div \cos ^2x + 2[\sin x \div \cos x] } / [1+2sinx cosx]\]
\[(1+ 2sinx cosx /\cos^2x ) / 1+2sinx cosx\]

1+2sinx cosx will be cancelled out which leaves
\[1/\cos^2x = \sec^2 x\]

Answer 7

well i did this if u cannot understand let me know will try better

Answer 8

well @kira you get it or not?

Answer 9

Assume they are equal, then multiply both sides by 1 + sin(2x)

\[\sec ^{2}x+2tanx=\sec ^{2}x+2(sinx)(cosx)(\sec ^{2}x)\]
\[2tanx=2(sinx)(cosx)(\sec ^{2}x)\]
\[2tanx=2(sinx)(cosx)(1/\cos^{2}x)\]
\[tanx=(sinx)(cosx)(1/\cos^{2}x)\]
\[tanx=((sinx)(cosx))/(\cos^{2}x)\]
\[tanx=(sinx)/(cosx)\]
Which we know is true.

Answer 10

ian what software did you write that on?

Answer 11

there is an equation button on the bottom.
I think it's LaTeX, you can type stuff without the equation button if you know the language. (I don't)

Answer 12

ahh.. just saw that thanks

Answer 13

\[1/\cos^2x+2sinx/cosx \div Sin^2x+\cos^2x+2sinxcosx\] (as Sin^2x6+Cos^2x=1)

[1+2cosxsinx/\cos^2x\]
\[Sin^2x+Cos^2x \div Cos^2x\]
\[1/Cos^2x\]
\[Sec^2X\]

Answer 14

Guys thanks...