Trigonometric proofs

Question

How do you solve

tan2x = 2/cotx-tanx

so far i've got

sin2x/cos2x = 2/(cos/sin-sin/cos)

sin2x/cos2x = 2/(cos^2xsin^2x/sinxcosx)

sin2x/??? = sin2x/2(cos2x)

 

how would i finish

Answer 1

It will be easier to start from the right hand side.

2 / (cot(x) - tan(x))
= 2sin(x)cos(x) / (sin(x)cos(x)(cot(x) - tan(x)))
= sin(2x) / (sin(x)cos(x)(cot(x) - tan(x))
= sin(2x) / (sin(x)(cos(x)cot(x) - cos(x)tan(x)))
= sin(2x) / (sin(x)(cos(x)cot(x) - sin(x)))
= sin(2x) / (sin(x)cos(x)cot(x) - sin(x)sin(x))
= sin(2x) / (cos(x)cos(x) - sin(x)sin(x))
= sin(2x) / (cos^2 (x) - sin^2 (x))
= sin(2x) / cos(2x)
= tan(2x)

Comments

we're taught to put everything into sins and cos' to simplify everything, and i'm very confused with how you did what you did on the right side, with the bottom of the fraction..

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I am bringing in sin(x) and cos(x) into cot(x) - tan(x).
Since tan(x) is sin(x)/cos(x), bringing in cos(x) will result in sin(x). cot(x) will not change, so we have cos(x)cot(x).
Similarly, since cot(x) is cos(x)/sin(x), bringing in sin(x) will result in cos(x). sin(x) will not change, so we will have sin(x)sin(x).

Answer 2

tan2x = 2/cotx-tanx                           tan2x = sin2x/cos2x

sin2x/cos2x =2/cotx - tanx             sin2x = 2sinxcosx , 

                                                          cos2x = cos^2x - sin^2x 
 

2sinxcosx

--------------------- = 2/cotx - tanx

cos^2x - sin^2x

 

2sinxcosx/sinxcosx

---------------------------------- = 2/cotx - tanx

(cos^2x - sin^2x)/sinxcosx

 

        2

--------------------------------------------------- = 2/cotx - tanx

cos^2x/sinxcosx - sin^2x/sinxcosx

              2

-------------------------------------------- =  2/cotx - tanx

           cosx/sinx - sinx/cosx

               2

-----------------------------------  =2/cotx - tanx

              cotx - sinx   

2/( cotx - tanx) = 2(/cotx - tanx)
 

Answer 3

tan 2x=2tanx/1-tan^2x. then divide by tan x in numerator n denominator . u ll get u answer