How do you simplify the expression #(sintheta+tantheta)/(1+sectheta)#?

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Answer 1 · 2016-09-09T14:30:36

#sin(theta)#

We have: #(sin(theta) + tan(theta)) / (1 + sec(theta))#

Let's apply two standard trigonometric identities; #tan(theta) = (sin(theta)) / (cos(theta))# and #sec(theta) = (1) / (cos(theta))#:

#= (sin(theta) + (sin(theta)) / (cos(theta))) / (1 + (1) / (cos(theta)))#

#= ((sin(theta)cos(theta) + sin(theta)) / (cos(theta))) / ((cos(theta) + 1) / (cos(theta)))#

#= (sin(theta)cos(theta) + sin(theta)) / (cos(theta) + 1)#

Let's apply the Pythagorean identity #cos^(2)(theta) + sin^(2)(theta) = 1#:

#= (sin(theta) (cos(theta) + 1)) / (cos(theta) + cos^(2)(theta) + sin^(2)(theta))#

We can rearrange the Pythagorean identity to get:

#=> sin^(2)(theta) = 1 - cos^(2)(theta)#

Let's apply this to get:

#= (sin(theta) (cos(theta) + 1)) / (cos(theta) + cos^(2)(theta) + 1 - cos^(2)(theta))#

#= (sin(theta) (cos(theta) + 1)) / (cos(theta) + 1)#

#= sin(theta)#