Question

Show that tan(a-b)+tanb/1-tan(a-b)tanb=tana
Please someone help me with this...

Answer 1

this needs rewriting. hard to tell what the fractions are

Answer 2

numerator is: tan(a-b)+tanb
denominator is: 1-tan(a-b)tanb
the whole thing equals to tana

Answer 3

\[\tan(a)=\frac{ \tan(a-b)+\tan(b) }{ 1-\tan(a-b)\tan(b) }\]

Answer 4

yes

Answer 5

Use Addition Formula: \(tan(a+b) = \frac{tana+tanb}{1-tana.tanb}\)

Answer 6

that's what I tried then I got:

Answer 7

\[\tan(a)=\tan(a+b)(1-\tan(a)\tan(b))-\tan(b)\]

Answer 8

\(\frac{tan(a-b) + tanb}{1-tan(a-b).tanb}=tan[(a-b)-b]=tana\)

Answer 9

how do you get it to equal \[\tan[(a-b)-b]=tana\]

Answer 10

tan[(a-b)+b]=tan a
because tan (a+b)= (tan a + tan b)/(1- tan a tan b)

Answer 11

but how do I get from the angle sum identity to the one with the double braquets?

Answer 12

OMG I got big mistake

Answer 13

Sorry @Smilesong123

Answer 14

\(tan(a-b)+b=tana\) :)

Answer 15

You just use Addition Formula

Answer 16

here we are using tan(a+b) formula and writing tan(a-b){ as in your question} as tan a
else you can use this formula \[\tan(a-b)= \frac{ \tan(a)- \tan(b }{ 1+ \tan(a)\tan(b)}\]
in your question and solve the equation, you will get the same answer

Answer 17

but how do you get that answer?

Answer 18

\[\frac{tan\ a+tan\ b}{1-tan\ a.tan\ b}=\frac{\frac{sin\ a}{cos\ a}+\frac{sin\ b}{cos\ b}}{1-\frac{sin\ a}{cos\ a}\frac{sin\ b}{cos\ b}}=\frac{\frac{sina\ cosb+cosa\ sinb}{cosa\ cosb}}{\frac{cosa\ cosb - sina\ sinb}{cosa\ cosb}}=\frac{sin(a+b)}{cos(a+b)}=tan(a+b)\]

Answer 19

\(let\ a(in\ formulas) = a-b\\ \quad
b(in\ formulas) = b\)

Answer 20

thank you so much! now it makes more sense! :D

Answer 21

^_^