If $tanA=-4/3$ where $pi/2 < A < pi$ and $cosB=5/6$ where $0 < B < pi/2$ , find $tan(A-B)$ ?

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Answer 1 · 2017-02-27T09:04:41

$tan(A-B)=-(432+125sqrt11)/49=-17.277$

As $tanA=-4/3$ and $pi/2 < A < pi$ , $/_A$ lies in Quadrant 2

As $cosB=5/6$ and $0 < B < pi/2$ , $/_B$ lies in Quadrant 2.

Hence $secB=1/(5/6)=6/5$

and $tanB=sqrt(sec^2B-1)=sqrt((6/5)^2-1)=sqrt(11/25)=sqrt11/5$ and it is positive as $/_B$ lies in Quadrant 2.

$:.tan(A-B)=(tanA-tanB)/(1+tanAtanB)$

= $(-4/3-sqrt11/5)/(1-4/3xxsqrt11/5)$

= $(-20-3sqrt11)/(15-4sqrt11)$
(multiplying numerator and denominator by $15$ )

Let us now rationalize the denominator by multiplying numerator and denominator by $15+4sqrt11$ , and we get

$tan(A-B)=(-20-3sqrt11)/(15-4sqrt11)xx(15+4sqrt11)/(15+4sqrt11)$

= $(-300-80sqrt11-45sqrt11-12xx11)/(15^2-(4sqrt11)^2)$

= $(-300-125sqrt11-132)/(225-176)$

= $-(432+125sqrt11)/49=-17.277$

If tanA=-4/3tanA=-4/3 where pi/2 < A < pipi/2 < A < pi and cosB=5/6cosB=5/6 where 0 < B < pi/20 < B < pi/2, find tan(A-B)tan(A-B)?