1/cos x-cos x/1+sinx=sin x/cos x?

Answer 1

I assume you mean:

1/cos x - cosx/(1+sin x) = sin x / cos x

Multiply through by cos x:

1 - cos^2 x/(1 + sin x) = sin x

Now, remember that cos^2 x + sin^2 x = 1... so that means cos^2 x = 1 - sin^2 x. You can factor 1-sin^2x as (1 - sin x) (1 + sin x); therefore, the (1 + sin x) will cancel out the denominator, leaving:

1 - (1 - sin x) = sin x

1 - 1 + sin x = sin x

sin x = sin x

... so the equation checks out as true, as long as cos x is not 0, and sin x is not -1, since those would make the equation invalid (zero denominators).

Answer 2

This is kind of a pain in the neck, but on the left side of the equation you'll need a common denominator, which will be cosx(1+sinx).

When you fix the numerators for the common denominator you'll get (1+sinx)/[cosx(1+sinx)] - cos^2x/[cosx(1+sinx)]

Then combine the two numerators. There aren't any common factors, but you have an identity that says sin^2x + cos^2x = 1. So you can rearrange it to replace the cos^2 with 1- sin^2.

Your numerator will now be 1+ sinx -(1-sin^2x) = sinx + sin^2x

Take out the common factor of sinx and you'll get
sinx(1+sinx)
The 1+sinx will cancel the 1+sinx in the denominator and leave
sinx/cosx.

I know this is probably confusing. It's fraction arithmetic with sin and cos. Where it says sin^2x I mean sin squared x, not sin 2x. I hope this helps.

Answer 3

you mean
(1/cos x) - cos x/(1+sinx)
to get rid of 1+sin x in denominator multiply by 1- sin x to make 1- sin ^2 x or cos^2 x. Becuase u multiply denominator same should apply to numerator

= (1/cos x) - cosx(1-sin x)/((1+sin x)(1-sin x))
= 1/cos x - cos x(1-sin x)/(1-sin^2 x)
= 1/cos x - cos x(1-sinx)/cos^ 2 x
= 1/cos x -(1-sin x)/cos x
= (1-1+sin x)/cos x
= sin x/cos x
= RHS
QED

Answer 4

1/cos x - cos x/(1+sin x) = sin x/cos x

Hopefully, this is the expression you intended.

Multiply first term by (1 + sin x)/(1 + sin x) and the second term by cos x/cos x

(1 + sin x)/[cos x(1 + sin x)] - cos^2 x/[cos x(1 + sin x)]

(1 + sin x - cos^2 x)/[cos x(1 + sin x)]

And cos^2 = 1 - sin^2 x

(1 + sin x - 1 + sin^2 x)/[cos x(1 + sin x)]
sin x(1 + sin x)/[cos x(1 + sinx)]
sin x/cos x

Answer 5

LHS =
( 1+sin x) - (cos x )^2 / (1 + sin x )cos x
( 1+sin x) - (1 - (sin x)^2 / (1 + sin x )cos x
( 1+sin x) - (1 - sin x)(1 + sin x) / (1 + sin x )cos x
(1 - 1 + sin x) / cos x
sin x / cos x
= RHS

Piece of cake. Would have been easier if given on paper.

Answer 6

cos x /a million-sin x -sin x/cos x locate lcm of denominators u get = (a million-sin x)(cos x) so u get cos^2 x -sin x + sin^2 x /(a million-sin x )(cos x) (a million-sin x)/(a million-sinx)(cos x) (a million-sin x ) gets canceled. for this reason we get a million/cos x if u have not understood i'm going to objective to function my working to a area wait..

Answer 7

Obviously it does.
(1/cos(x)) - (cos(x)/(1+sin(x)) = sin(x)/cos(x)
1 - cos²(x)/(1+sin(x)) = sin(x)
1+sin(x) -cos²(x) = sin(x) + sin²(x)
1 = sin²(x) + cos²(x)
How much easier does it get? ☺


Doug

Answer 8

1/cosx -cosx/1+sinx)=!+sinx-cos^2x/cosx(1+sinx)
=sinx+sin^2x/cosx(1+sinx)
sinx(1+sinx)/cosx(1+sinx)
=sinx/cosx
hence proved

Answer 9

this question needs rephrasing with maybe some brackets to show what is being divided by what, ok