Prove 1/(cosx + sinxtanx) = cosx?

Question

I don't know about arcs or pi yet!

Answer 1

1/(cosx+sinxtanx)
=1/(cosx+sinx*sinx/cosx)
=1/(cos^2x+sin^2X)/cosx
=cosx/(cos^2x+sin^2x)
=cosx as cos^2x+sin^2x=1

Answer 2

It seems very strange that you're proving trig identities if you haven't heard of "pi", but fortunately you don't need it for this proof.

You do need to know that tanx = sinx/cosx, and you need the single most important trigonometric identity: (cosx)^2+(sinx)^2=1.

What you need to do is replace things with what they're equal to and then keep simplifying.

1/(cosx + sinxtanx) =
1/(cosx + sinx*(sinx/cosx)) =
1/(cosx + (sinx)^2/cosx)) =
1/(cosx + ((1 - (cosx)^2)/cosx))

Now multiply the top and bottom of the fraction by cosx

cosx*1 / cosx(cosx + ((1 - (cosx)^2)/cosx)) =
cosx / ((cosx)^2 + 1 - (cosx)^2) =
cosx / 1 =
cosx

Answer 3

1/(cosx +sinx*sinx/cosx)
1(cos^2x/cosx +sin^2x/cosx)
1/(cos^2x+sin^2x)/cosx)
1/1/cosx
1*cosx

start by changint tan to sin/cos...get LCD for denominator
use identity that cos sqared + sin squared =1

Answer 4

how come u dunno pi, and yet ur alredy on trigo? neway

1/(cosx + sinx.tanx)

= 1/(cosx + sinx.(sinx/cosx))

= 1
-------------------
cosx^2 + sinx^2
-------------
cosx

= cosx

Hence proved!

The identities used are:
tanx = sinx/cosx
sinx^2 = sinx squared
same for cosx^2
sinx^2 + cosx^2 = 1 (accepted truth!)
1/1/a = a