Question

If `tan(a+b)=1 and tan(a-b)=1/7` then how will you find out the values of `tana and tanb`?

Answer 1

`tana=-2, 1/2; & tanb=-3,1/3.`

Explanation:

METHOD I

Let `tana=x, tanb=y.`

Given that,`tan(a+b)=1rArr(tana+tanb)/(1-tana tanb)=1rArrx+y=1-xy...(1)`

Similarly,`tan(a-b)=1/7rArrx-y=1/7(1+xy)..........(2)`

`(1)+(2) rArr 2x=8/7-6/7xy,`or, `x=4/7-3/7xy,`i.e.,`x(1+3/7y)=4/7,`giving, `x=4/(7+3y).`

We submit this `x` in `(1)` to see that.
`4/(7+3y)+y+(4/(7+3y))y=1,` or,
`4+7y+3y^2+4y=7+3y,` i.e., `3y^2+8y-3=0.`

Hence, `y=tanb=-3,1/3.`

Using `x=4/(7+3y),` we get, `x=tana=-2,1/2.`

Answer 2

`tana=-2, 1/2 ; tanb=-3, 1/3.`

Explanation:

METHOD II:-

Take, `a+b=C, a-b=D,` then, by what is given, `tanC=1, tanD=1/7...(1)`

Observe that, `C+D=2a.`
`C+D=2a.rArrtan(C+D)=tan2a.rArr(tanC+tanD)/(1-tanCtanD)=tan2a`

Here, we use `(1)` to get, `(1+1/7)/(1-1/7)=tan2a,` or, `tan2a=4/3..(2)`

Recall that `tan2a=(2tana)/(1-tan^2a)........(3).` so, if, `tana=x, (2) & (3) rArr(2x)/(1-x^2)=4/3,rArr3x=2-2x^2,rArr2x^2+3x-2=0,rArr(x+2)(2x-1)=0,rArrx=tana=-2, 1/2,.` as in METHOD I!

`tanb` can similarly be obtained using `C-D=2b.`