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

2 Answers
Ratnaker Mehta
Jul 1, 2016

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.

Ratnaker Mehta
Jul 1, 2016

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.

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