Question

tanA+tanB/1-tanA.tanB=sin(a+b)/cos(a+b)

Answer 1

You might wanna use sum to product formula.

Answer 2

Proving identities?

Answer 3

hy sam

Answer 4

can u solve my optional math question

Answer 5

Let's start with the right-hand side
\[\Large \frac{\sin(A+B)}{\cos(A+B)}\]

Now hopefully, recalling the sum identitity
\[\Large = \frac{\sin(A)\cos(B) + \cos(A)\sin(B)}{\cos(A)\cos(B) - \sin(A)\sin(B)}\]

Answer 6

You there, @bijay56 ?
Catch me so far?

Answer 7

hy terenzreignz

Answer 8

You can call me Terence :)
More importantly, do you understand the expansion?

Answer 9

yes i uderstood and after that

Answer 10

After that, a tricky step... we'll multiply 1 to the expression... but not just any "1" but a special variant of 1...
\[\Large = \frac{\sin(A)\cos(B) + \cos(A)\sin(B)}{\cos(A)\cos(B) - \sin(A)\sin(B)}\times \frac{\frac{1}{\cos(A)\cos(B)}}{\frac1{\cos(A)\cos(B)}}\]

Answer 11

So, going ahead and distributing...
\[\Large = \frac{\frac{\sin(A)\cos(B)}{\cos(A)\cos(B)} + \frac{\cos(A)\sin(B)}{\cos(A)\cos(B)}}{\frac{\cos(A)\cos(B)}{\cos(A)\cos(B)} -\frac{ \sin(A)\sin(B)}{\cos(A)\cos(B)}}\]

Answer 12

The fun part... I'll leave this to you, but cancel out what can be cancelled, and remember that...
\[\huge \frac{\sin(x)}{\cos(x)}=\tan(x)\]

Enjoy the slashing ;)

Answer 13

thanks terence

Answer 14

No problem :)

Answer 15

i have some work i will come in minute ternce

Answer 16

\[\huge = \frac{\frac{\sin(A)\cos(B)}{\cos(A)\cos(B)} + \frac{\cos(A)\sin(B)}{\cos(A)\cos(B)}}{\frac{\cos(A)\cos(B)}{\cos(A)\cos(B)} -\frac{ \sin(A)\sin(B)}{\cos(A)\cos(B)}}\]
this is almost done, by the way, just some slashing and converting, and you'll be able to turn this into the left-side in a jiffy.

Answer 17

hy

Answer 18

Yes?

Answer 19

cos(a+b).cos(a-b)=cos^2A-sin^2B

Answer 20

Expand the left-hand side.

Answer 21

how

Answer 22

This rule...
\[\large \cos(\alpha+\beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\]

Answer 23

ok and after that

Answer 24

Expand the left-hand side first, then I'll tell you the next step.

Answer 25

i have done now

Answer 26

What did you get?

Answer 27

cosA.cosB-sinA.sinB.cosA.cosB+sinA.sinB

Answer 28

Okay. But put the parentheses.
(cosAcosB - sinAsinB)(cosAcosB + sinAsinB)

Notice that it's of the form...
\[\large (x-y)(x+y)\]which is equal to...
\[\large x^2 - y ^2\]

Answer 29

now what to do

Answer 30

terence now what to do after that

Answer 31

What to do?
I told you already...
\[\Large [\cos(A)\cos(B) -\sin(A)\sin(B)][\cos(A)\cos(B) + \sin(A)\sin(B)]\]

It's just like this...
\[\Large (x-y)(x+y) = x^2 - y^2\]

Answer 32

Sorry, got cropped...\[\large [\cos(A)\cos(B) -\sin(A)\sin(B)][\cos(A)\cos(B) + \sin(A)\sin(B)]\]

Answer 33

It's like...
\[\large x = \cos(A)\cos(B) \\\large y = \sin(A)\sin(B)\]

Answer 34

now next step

Answer 35

You know me :P
No next step until you show me your progress :D

Answer 36

next step:cos^2A.cos^2b-sin^2A,sin^2B

Answer 37

next step:cos^2A.cos^2B-sin^2A.sin^2B

Answer 38

next step right or wrong

Answer 39

And from here, it's just algebraic manipulation.

Answer 40

Whoops, sorry....
\[\large \cos^2(A)\cos^2(B)- \color{green}{\sin^2(A)}\sin^2(B)\]


replace with...\[\large \cos^2(A)\cos^2(B)- \color{red}{[1-\cos^2(A)]}\sin^2(B)\]

Answer 41

If it makes it simpler, it may be tweaked into this...
\[\large \cos^2(A)\cos^2(B) \color{blue}{+[\cos^2(A)-1]}\sin^2(B)\]

Answer 42

hy

Answer 43

terence can u solve me another opt math problem