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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√2 but i really want to know how you actually solve it.

There's another one i got stuck on aswell:-

√3cos15° - sin15°

I was thinking about using sin(45-30) or something. Please help!

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

There's another one i got stuck on aswell:-

√3cos15° - sin15°

I was thinking about using sin(45-30) or something. Please help!

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√2 but i really want to know how you actually solve it.

There's another one i got stuck on aswell:-

√3cos15° - sin15°

I was thinking about using sin(45-30) or something. Please help!

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

There's another one i got stuck on aswell:-

√3cos15° - sin15°

I was thinking about using sin(45-30) or something. Please help!

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=√3, and BD = 2, since ABD is isosceles,

Tan ADM=1/(2+√3)

working on from there, I get 1/√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

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.

*bobo*

the answer should be tan 30, which equals (1/√3)=(√3/3)

(1-tan15)/(1+tan15)

=(tan180-tan15)/(tan180+tan15)

=(tan165)/(tan195)

=tan-30

=1/√3

=√3/3

(1-tan15)/(1+tan15)

=(tan180-tan15)/(tan180+tan15)

=(tan165)/(tan195)

=tan-30

=1/√3

=√3/3

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

*bobo*

(1-tan15)/(1+tan15) does not equal √2

the answer should be tan 30, which equals (1/√3)=(√3/3)

(1-tan15)/(1+tan15)

=(tan180-tan15)/(tan180+tan15)

=(tan165)/(tan195)

=tan-30

=1/√3

=√3/3

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

the answer should be tan 30, which equals (1/√3)=(√3/3)

(1-tan15)/(1+tan15)

=(tan180-tan15)/(tan180+tan15)

=(tan165)/(tan195)

=tan-30

=1/√3

=√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

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 √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 √3cos15° - sin15° But thanx for your help! Is √3cos15° - sin15° anyway linked to what we've just solved?

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