Question

Hey guise
FAN AND MEDAL!!!!!!!

tanA+cosAsecBsecC=tanB+cosBsecAsecC and that it is also equal to tanC+cosCsecAsecB
thanks

Answer 1

@ikram002p

Answer 2

@mathmate

Answer 3

let me see...

Answer 4

I would try the classic!

Answer 5

what would that be? sorry im just a beginner

Answer 6

not so sure any more about the classic.... decompose everything into sin and cos

Answer 7

ok ill try that gimme a sec

Answer 8

except that here we have three different angles, A, B and C, so they don't cancel that fast.

Answer 9

You probably know something about what's relating A, B and C.
Would A+B+C=180 degrees or something like that?

Answer 10

wait sorry it says A+B+C=90

Answer 11

By the symmetry of it, it sounds more like A=B=C.

Answer 12

it says A+B+C=90 lol

Answer 13

ok, at least there's something to go on!
Have you done addition formulae?

Answer 14

ya

Answer 15

Wow, that's would be a big help.
How do you think it could help?

Answer 16

wait by addition formula do you mean cosx+cosy etc?

Answer 17

more like tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))
or something similar.

Answer 18

yea i've got that but not sure how it helps

Answer 19

Well, you could try eliminating A using A=90-B-C, and using addition formula to expand it, and hopefully something will cancel.

Answer 20

but how is that meant to be an expansion?

Answer 21

\[\cot(A+B) + (\sin(B+C)) \div cosBcosC\] thats what i have

Answer 22

something like that, but we have to be careful which ones to choose.

Answer 23

Perhaps we need to eliminate one of the variables A, B or C so we end up with only two variables.

Answer 24

uh a few steps in, we get tan b and cosBsinCsecC but the expansion of the tangent value sorta messed it up

Answer 25

I am still bugged by the symmetry of the three expression, which seems to tell me that I am missing something.

Answer 26

yea idk either :(

Answer 27

I will work on this on my own.
In the mean time you can try gurus like @ganeshie8.
Sorry! :(

Answer 28

thats ok

Answer 29

@ganeshie8 any ideas?

Answer 30

@No.name idea?

Answer 31

nopes sry

Answer 32

thats ok

Answer 33

@SolomonZelman

Answer 34

anyone?

Answer 35

@Purplerainbowcherry

Answer 36

@Mimi_x3

Answer 37

halp pls

Answer 38

sorry :( I literally suck at these kind of questions :/

Answer 39

@CrashOnce I cooked up something for you!

To show:
\(\large tanA+cosA secB secC = tanB+cosB seC secA = tanC+cosC secA secB\)

To do this, it would be less confusing to prove two simple relations ahead of time to simplify the proof.
These two proofs (called lemmas) require the use of addition/subtraction trigonometric formulas:
\(\large cos(B+C) = cosB cosC - sinB sinC\)
from which
\(\large cosB cosC = cos(B+C) + sinB sinC\)....(1)

From A+B+C=90, and
\(\large cos(X-Y) = cosX cosY + sinX sinY\)
We have
\(\large cos(B+C)=cos(90-A) = cos90 cosA + sin90 sinA\)
since cos90=0, sin90=1
\(\large cos(B+C)= sinA\) ....(2)

Starting with the first expression:
\(\large tanA + cosA secB secC\)
We reduce to sin and cos to help simplify things
\(\large \frac{sinA}{cosA} + \frac{cosA}{cosB cosC}\)
join expressions using common denominator:
\(\large \frac{sinA cosB cosC + cos^2A}{cosA cosB cosC}\)
Apply relation (1) above to cosB cosC:
\(\large \frac{sinA(cos(B+C) + sinB sinC) ) + cos^2A}{cosA cosB cosC}\)
Apply relation (2) above to cos(B+C):
\(\large \frac{sinA(sinA + sinB sinC) + cos^2A}{cosA cosB cosC}\)
Expanding:
\(\large \frac{sin^2A + sinA sinB sinC + cos^2A}{cosA cosB cosC}\)
\(\large \frac{sin^2A + cos^2A + sinA sinB sinC}{cosA cosB cosC}\)
Using \(sin^2+cos^A = 1\)
\(\large \frac{1+sinA sinB sinC}{cosA cosB cosC}\)...(3)

Hence
\(\large tanA+cosA secB secC \equiv \frac{1+sinA sinB sinC}{cosA cosB cosC}\) ...(4)

We can similarly show that
\(\large tanB+cosB secC secA \equiv \frac{1+sinB sinC sinA}{cosB cosC cosA}\) ...(5)

and
\(\large tanC+cosC secA secB \equiv \frac{1+sinC sinA sinB}{cosC cosA cosB}\) ...(6)

Since (4),(5),(6) all equal to the same expression symmetric in A,B and C, the given identity is proved.

Answer 40

ah thanks @mathmate

Answer 41

You're welcome! :)