IIT JEE , AIEEE Forum - Mathematics , Trignometry

varun.tinkle

1)

tan a +tan b=A

cot a +cot b= B

prove-

cot(a+b)=1/a-1/b

2)

sin(a+b)=1 sin(a-b)=1/2 where 0<a , b<90

find tan(a+2b) answer -root 3

and tan(b+2a) ans-1/root3

3)

tan x+tan(x+60)+tan(x+120)=3

prove that

3 tan x -tan^3 x =1

1-3 tan^2 x

pls expln in detail

rates assured

1-3tan^2x

ramkumar_november

1)

$\frac{1}{cota}+\frac{1}{cotb}=A$

$\frac{cota+cotb}{cota.cotb}=A$

$\frac{1}{A}=\frac{cota.cotb}{cota+cotb}$

$\frac{1}{B}=\frac{1}{cota+cotb}$

therefore

$\frac{1}{A}-\frac{1}{B}=\frac{cota.cotb-1}{cota+cotb}$ = cot(a+b)

LAMPARD

a+b=90
a-b=30
Solve to get a and b as 60 and 30.
Then find a+2b and find its tan which is tan120=-root(3)
tan(b+2a)=tan(150)=-1/root(3)

ramkumar_november

3)

we use this formula:

from the given equation we find that tan3x=1

so $\frac{3tanx-tan^3x}{1-3tan^2x}=1$

risin

a+b=pi/2,

a-b=pi/6

a=pi/3,b=pi/6,

Therefore,

tan(a+2b)=-root3,

tan(b+2a)=tan(pi/6+2pi/3)=-i/root3..

mathwiz

how did u get a+b as pi/2 and a-b as pi/6??

RyuAmakusa

@mathwiz 0 a+b = 90.

now0< a-b <90 and sin(a-b) = 1/2 so a-b = 30.

any clarifications.