What is the value of #sintheta+ costheta/sintheta - costheta# OR #(sin theta + costheta)/(sin theta - costheta)# if #sintheta = 3/5#?

2 Answers
EZ as pi
Aug 18, 2017

#17/15 = 1 2/15#

Explanation:

#sin theta = 3/5# tells us that two sides in a right-angled triangle are #3 and 5#. Using the Pythagorean triple #3:4:5# means that we have:

opposite side = 3, adjacent side = 4 and hypotenuse = 5

#sin theta =3/5#

#cos theta = 4/5#

#tan theta = 3/4#

#sin theta+ cos theta÷ sin theta- cos theta#

#= 3/5 + 4/5 div 3/5 - 4/5#

#= 3/5 +4/cancel5 xx cancel5/3 -4/5#

#3/5+4/3-4/5" "larr# find a common denominator

#(9+20-12)/15#

#=17/15#

#=1 2/15#

Noah G
Aug 18, 2017

#(sin theta + costheta)/(sintheta - costheta) = -7#

Explanation:

Here's another way of interpreting the question. If #sintheta = 3/5#, then find the value of #(sin theta + costheta)/(sin theta - costheta)#.

If we multiply both the numerator and the denominator by #sintheta + costheta#, we get:

#= (sin theta + costheta)/(sin theta - costheta) * (sin theta + costheta)/(sintheta + costheta)#

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

Use #sin^2x + cos^2x = 1#.

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

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

Now use #sin(2x) = 2sinxcosx#.

#=(1 + sin2x)/(2sin^2theta - 1)#

So now we have that if #sintheta = 3/5#, then #costheta = 4/5# (as derived by EZ as pi). So #sin(2x) = 2(4/5)(3/5) = 24/25#

#=(1 + 24/25)/(2(3/5)^2 - 1)#

#= (49/25)/(18/25 - 1)#

#= (49/25)/(-7/25)#

#= -7#

Hopefully this helps!

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