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

Jul 1, 2016

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

#### Explanation:

METHOD I

Let $\tan a = x , \tan b = y .$

Given that,$\tan \left(a + b\right) = 1 \Rightarrow \frac{\tan a + \tan b}{1 - \tan a \tan b} = 1 \Rightarrow x + y = 1 - x y \ldots \left(1\right)$

Similarly,$\tan \left(a - b\right) = \frac{1}{7} \Rightarrow x - y = \frac{1}{7} \left(1 + x y\right) \ldots \ldots \ldots . \left(2\right)$

$\left(1\right) + \left(2\right) \Rightarrow 2 x = \frac{8}{7} - \frac{6}{7} x y ,$or, $x = \frac{4}{7} - \frac{3}{7} x y ,$i.e.,$x \left(1 + \frac{3}{7} y\right) = \frac{4}{7} ,$giving, $x = \frac{4}{7 + 3 y} .$

We submit this $x$ in $\left(1\right)$ to see that.
$\frac{4}{7 + 3 y} + y + \left(\frac{4}{7 + 3 y}\right) y = 1 ,$ or,
$4 + 7 y + 3 {y}^{2} + 4 y = 7 + 3 y ,$ i.e., $3 {y}^{2} + 8 y - 3 = 0.$

Hence, $y = \tan b = - 3 , \frac{1}{3.}$

Using $x = \frac{4}{7 + 3 y} ,$ we get, $x = \tan a = - 2 , \frac{1}{2.}$

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, $\tan C = 1 , \tan D = \frac{1}{7.} . . \left(1\right)$

Observe that, $C + D = 2 a .$
$C + D = 2 a . \Rightarrow \tan \left(C + D\right) = \tan 2 a . \Rightarrow \frac{\tan C + \tan D}{1 - \tan C \tan D} = \tan 2 a$

Here, we use $\left(1\right)$ to get, $\frac{1 + \frac{1}{7}}{1 - \frac{1}{7}} = \tan 2 a ,$ or, $\tan 2 a = \frac{4}{3.} . \left(2\right)$

Recall that $\tan 2 a = \frac{2 \tan a}{1 - {\tan}^{2} a} \ldots \ldots . . \left(3\right) .$ 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!

$\tan b$ can similarly be obtained using $C - D = 2 b .$