Prove tan θ / sin θ = cos θ/ 1 – sin^2 θ?

Question

Prove tan θ / sin θ = cos θ/ 1 – sin^2θ

Answer 1

trig identity: sin^2☻ = 1-cos^2☻
therefore cos^2☻=1-sin^2☻

= cos☻/cos^2☻
=1/cos☻
tan☻= son☻/cos☻
therefore
tan/sin= sin/cos/sin= 1/cos
therefore they r both equal to each other

which all equals sec☻

Answer 2

Proving something is true by saying, "just plug any value in and you'll notice that it's never false" is the wrong way to approach things, specifically because there are an infinite amount of numbers. You have to prove this like you would prove all identities.

The first step would be to choose the more complex side. I'm going to choose the left hand side, for

tan(t) / sin(t) = cos(t) / [1 - sin^2(t)]

LHS = tan(t) / sin(t)

Change everything to sines and cosines.

LHS = [sin(t)/cos(t)] / sin(t)

Get rid of the complex fraction by multiplying top and bottom by cos(t).

LHS = sin(t) / [sin(t)cos(t)]

Note that sin(t) on the top and bottom cancel.

LHS = 1 / cos(t)

Let's stop here and work with the right hand side.

RHS = cos(t) / (1 -sin^2(t))

Note that since cos^2(t) = 1 - sin^2(t) by the famous trig identity. For that reason,

RHS = cos(t) / cos^2(t)

We can reduce this by canceling cos(t) on the top with the bottom, to get

RHS = 1 / cos(t)

Note now that LHS = RHS. Thus we're done with proving our identity.

Answer 3

use the identities:

tanθ = sinθ / cosθ

cos^2 θ + sin^2 θ = 1 --> cos^2 θ = 1 - sin^2 θ

plug in the identities, you get

(sin θ/cos θ)/sinθ = cos θ / cos^2 θ

cancel the common factors, and you get

1/ cos θ = 1/ cos θ ., done

Answer 4

Prove

tan θ / sin θ = cos θ/ (1 – sin² θ)

Let's work with the left hand side.

tan θ / sin θ = (sin θ / cos θ) / sin θ = 1 / cosθ
= cos θ / cos² θ = cos θ / (1 - sin² θ)

qed

Answer 5

L.H.S.-
tan θ/sin θ
= sinθ/cosθsinθ
sinθ and sinθ cancels
=1/cosθ
multiply numerator and denominator by cosθ
=1*cosθ/cosθ*cosθ
=cosθ/cos^2θ
=cosθ/1-sin^2θ (cos^2θ=1-sin^2θ)
=R.H.S.

Answer 6

1- sin ^2 t = cos ^2 t

RHS - cos t/(1- sin ^2 t) = cos t/ cos^2 t = 1/ cos t = tan t/(tan t cos t) = tan t/ sint
= LHS
as sin t/ cos t

Answer 7

(sin θ/cos θ)/sin θ
cos θ
1/cos^2θ(cos θ)
cos θ/cos^2θ
cos θ/1-sin^2θ

Answer 8

These trig identies fall out by using imaginary numbers. plug in the imaginary forms foer each trig function and reduce

Answer 9

I'm guessing you want this in proof form?

tanx/sinx =
sinx/(cosxsinx) =
1/cosx =
cosx/(cosxcosx) [multiply numerator and denom. by cosx] =
cosx / (cos^2x)
cosx / (1-sin^x) = cosx / (1-sin^2x)

heh. that was actually kinda fun.

Answer 10

tan A=sinA/cosA

therefore
tan A/sinA= sinA/sinA*cosA=!/cosA


solvin the right side
1-sin^2A=cos^2A....since sin^2a+cos^2A=1

cosA/cos^2A=1/cosA
hence it is proved