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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