Trig Idenity, HELP PLEASE?

Question

This trig identity is tough (well 4 me anyways). Could you please help? What are the steps in solving this?

PROVE the identity. Work only one side of the equation.

sin2xsecx =2tanxcosx

Answer 1

sin(2x)sec(x) = 2tan(x)cos(x)

The more complex side is the left side (in my opinion); I disagree with you. The complex side is always the one you work with, to get to the simple side.

LHS = sin(2x)sec(x)

Using the double angle identity sin(2x) = 2sin(x)cos(x),

LHS = 2sin(x)cos(x)sec(x)

Change everything to sines and cosines.

LHS = 2sin(x)cos(x)[1/cos(x)]

Note the cancellation between cos(x) and 1/cos(x).

LHS = 2sin(x)

I'm going to do something funny here; I'm going to put this as a fraction over 1.

LHS = 2sin(x)/1

Now, I'm going to multiply top and bottom by cos(x); this doesn't change anything because cos(x)/cos(x) = 1.

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

I'm going to now split this up as a product of trig functions.

LHS = 2[sin(x)/cos(x)][cos(x)/1]

By definition

LHS = 2tan(x)cos(x) = RHS

Answer 2

Taking the RHS , we get
2 tanx cosx = 2 (sinx / cosx) * cosx = 2 sinx cosx * 1/ cosx

Now 2 sinx cosx = sin2x and 1/ cosx = secx

Therefore , sin2x cosx.

Answer 3

We start off from the Left Hand Side(LHS)...
LHS= sin2xsecx
= 2(sinx)(cosx)(secx) [ remember sin2x= 2(sinx)(cosx)]
= 2(sinx)(secx)(cosx) [ tanx=(sinx)/(cosx) = (sinx)(secx)]
= 2(tanx)(cosx)
= RHS

Answer 4

(sin2x)secx=(2sinxcosx) secx = 2 sinx (cosx secx)=2 sinx secx cosx = (2 sinx secx) cosx = (2 sinx/cosx) cosx = 2 tanx cosx