Question

Prove the identity 1+sinx/cosx+cosx/1+sinx=2/cosx

Answer 1

|sinx^2 + cosx^2 = 1

1+ 2sinx + 1 = 2 + 2sinx

2 + 2sinx = 2 + 2sinx

Answer 2

Sorry hershey we should prove that the left hand side is identical to the right hand side

Answer 3

Get common denominators on the left then add
The LCD is cos x(1 + sin x) so you get
(1 + sin x)(1 + sin x) / [cos x (1 + sin x)] + cos x (cos x) / [cos x (1 + sin x)]
= 1 + 2 sin x + sin^2 x + cos^2 x on top and since sin^2 + cos^2 = 1, this is
2 + 2 sin x or
2(1 + sin x) over [cos x (1 + sin x)] and you cancel the (1 + sin x) leaving what you want

Answer 4

I can give you

1.Results
2.Solution
3.Input
4. Atl form (Alternate)
5. Atl form assuming x is real:
6. Number line

Lol :D

Answer 5

\[(1+sinx)/cosx + cosx/(1+sinx)=2/cosx\]
\[cosx/(1+sinx)=2/cosx- (1+sinx)/cosx\]
\[cosx/(1+sinx)=(2-1-sinx)/cosx\]
\[\cos ^{2}x=(1+sinx)(1-sinx)\]
\[\cos ^{2}x=1-\sin ^{2}x\]
\[\sin ^{2}x+\cos ^{2}x=1\]
\[1=1\]
Therefore, LHS=RHS.
Hope this helps.

Answer 6

Well thanks a lot guys vengeance kisses method is right and correct boss we end up having2/cosx which is on the ryt hand side :)

Answer 7

So as long you get the right the right answer, I don't care who gives it :)

Answer 8

thats more like it :)