prove:cosx+sinxtanx=2?

Question

sorry, its supposed to be solve cosx+sinxtanx=2 not prove it. Sorry for the mistake.

Answer 1

OK, so if it's an EQUATION, then:

tanx = sinx/cosx, so

cosx + sinx(sinx/cosx) = 2
cosx + sin²x/cosx = 2
(cos²x + sin²x)/cosx = 2
1/cosx = 2
cosx = 1/2

The cosine is 1/2 at 60° (π/3) and -60° (5π/3).

So really the most general answer is:

π/3 ± 2kπ or 5π/3 ± 2kπ, where k is any integer.

Answer 2

this is an equation. the solution is :
cosx + sinxtanx = 2 <=> . . .
1/cosx = 2 <=>
1/2 = cosx <=>
cos (pi/3) = cosx <=>
x = 2k(pi) +- pi/3 (k in Z)

Answer 3

First turn Tanx into sinx/cosx:
cosx + sinx(sinx/cosx)=2

Then multiply:
cosx + sin^2x/cosx=2

Make a common denominator such as cosx and combine:
cosx(cosx/cosx) +sin^2x/cosx=2
cos^2x/cosx +sin^2x/cosx=2
(cos^2x+sin^2x)/cosx =2

You know that cos^2x+sin^2x =1 so plug that in:
1/cosx = 2

And 1/cosx = secx plug that in also:
secx = 2

(Maybe your problem is wrong because secx doesn't equal 2. Make sure you typed the problem in right.)

Answer 4

You have mis-typed the identity. Since, tan(x) = sin(x)/cos(x) your "identity" becomes

cos(x) + sin(x)(sin(x)/cos(x)) = 2

Let, x = 0

Now you have

cos(0) + sin(0)(sin(0)/cos(0)) = 2 or 1 + 0 = 2 or 1 =2, which seems incorrect to me. :)

HTH

Charles

Answer 5

sin(x)tan(x) = sin(x)sin(x)/cos(x)
= sin(x)^2/cos(x) = (1-cos(x)^2)/cos(x)
= 1/cos(x) - cos(x)
then, cos(x) + sin(x)tan(x) = cos(x) + 1/cos(x) - cos(x) = 1/cos(x)
??? not 2