Rewriting trigonometric identities

Question

How do I rewrite (tanθ+secθ)/cosθ in terms of sinθ?

(fully simplified)

Answer 1

Hi Gnarls,

 

Since tan = sin/cos and sec = 1/cos you can write:

 

(sinx/cosx) + (1/cosx)

--------------------------

          cosx

 

Think of this as a complex fraction which can be simplified by multiplying top and bottom of the "big" fraction by cos x.

 

That results in :

 

(sinx + 1)

------------

   (cos²x)

 

Replace the denominator by 1 - sin²x and change the order of the terms in the numerator.

 

(1 + sin x)

--------------

  1 - sin²x

 

Finally the denominator is a difference of squares so can be written as (1+sinx)(1-sinx). This results in the (1+sinx) factor in numerator and denominator cancelling out.

 

The result: 1/(1-sinx)

 

Here is some proof:

 

https://www.desmos.com/calculator/3qtyexkalx

 

The graphs of both f and g are identical.

 

 

Answer 2

(tanθ + secθ)/cosθ = (sinθ/cosθ + 1/cosθ)/(cosθ)

 

                             = (sinθ + 1)/cos²θ

 

                             = (sinθ + 1) / (1 - sin²θ)

 

                             = (sinθ + 1)/[(1 - sinθ)(1 + sinθ)]

 

                              = 1/ (1 - sinθ)

 

Answer 3

Let T be the angle Theta

 

The numerator is:

 sin T/cos T + 1/cos T  = (sin T + 1)/cos T

 

since tangent is sine over cosine and secant is 1 over cosine.

 

 

 

Dividing by cos T, ( sin T + 1 )/(cosT)^2 (sin T + 1)/(1 - (sinT)^2) (sin T + 1)/ (1 + sin T)(1- sin T) 1/(1 - sin T)