# Trigonometric identities

Hi! This looks simple but I'm not sure where I'm going wrong, I've used the identity tan(A-B)= (tanA-tanB)/(1+tanAtanB) but I end up with an odd answer. Here's the question:-

solve (1-tan°15)÷(1+tan°15) without using a calculator. I know the answer is&#8730;2 but i really want to know how you actually solve it.

There's another one i got stuck on aswell:-
&#8730;3cos15° - sin15°
Da Mouse
Hi! This looks simple but I'm not sure where I'm going wrong, I've used the identity tan(A-B)= (tanA-tanB)/(1+tanAtanB) but I end up with an odd answer. Here's the question:-

solve (1-tan°15)÷(1+tan°15) without using a calculator. I know the answer is&#8730;2 but i really want to know how you actually solve it.

There's another one i got stuck on aswell:-
&#8730;3cos15° - sin15°

To see tan 15°

draw an equilateral triangle ABC, and drop a perpendicular from B to the middle of AC. (M)

Produce MB to D, so that AB=BD. Now ADM = 15°

Let AB=2, so AM = 1, so BM=&#8730;3, and BD = 2, since ABD is isosceles,
working on from there, I get 1/&#8730;3 as the (eventual) answer...

You can get Tan 75° similarly. Maybe interesting, but not the quickest route, I'm sure...

I must now check the above...

Aitch
(1-tan15)/(1+tan15) = (tan90-tan15)/(1+tan15) = tan(90-15) = tan75 = tan(30+45) then expand
ssmoose
(1-tan15)/(1+tan15) = (tan90-tan15)/(1+tan15) = tan(90-15) = tan75 = tan(30+45) then expand

Nice method, but not good to replace 1 with tan 90.
(1-tan15)/(1+tan15) does not equal &#8730;2

the answer should be tan 30, which equals (1/&#8730;3)=(&#8730;3/3)

(1-tan15)/(1+tan15)
=(tan180-tan15)/(tan180+tan15)
=(tan165)/(tan195)
=tan-30
=1/&#8730;3
=&#8730;3/3

i then checked my answer on a calc and im pretty sure its correct
Nice method, but not good to replace 1 with tan 90.

Ah yes, haven;t done this since June so mistakes will be made. Replace all 90s in my solution above with 45s *hopes that is the right value this time*
*bobo*
the answer should be tan 30, which equals (1/&#8730;3)=(&#8730;3/3)

(1-tan15)/(1+tan15)
=(tan180-tan15)/(tan180+tan15)
=(tan165)/(tan195)
=tan-30
=1/&#8730;3
=&#8730;3/3

Not a good idea to replace 1 with tan 180 since tan 180=0.
evariste
Not a good idea to replace 1 with tan 180 since tan 180=0.

oops!
(1-tan15)/(1+tan15)

(1-tan15)/(1+tan15)
=(tan45-tan15)/(tan45+tan15)
= tan30 =1/&#8730;3
*bobo*
(1-tan15)/(1+tan15) does not equal &#8730;2

the answer should be tan 30, which equals (1/&#8730;3)=(&#8730;3/3)

(1-tan15)/(1+tan15)
=(tan180-tan15)/(tan180+tan15)
=(tan165)/(tan195)
=tan-30
=1/&#8730;3
=&#8730;3/3

i then checked my answer on a calc and im pretty sure its correct

I agree with that answer. (See my post above)

Aitch
Aitch
I agree with that. (See my post above)

Aitch

sorry, didn't see that
then my text book is wrong! bl**dy heineman lol Its a new textbook you see and my teacher says that they expect teachers to spot the mistakes and try to contact them. I was getting to an answer like that i think but i kept thinking that the answer was &#8730;2 - and I saw that my calculations was heading to something different - sorry I have a habit of rambling on but there was one thing..... I still cant get to how to solve &#8730;3cos15° - sin15° But thanx for your help! Is &#8730;3cos15° - sin15° anyway linked to what we've just solved?