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# if tan(a-b)=1/4, tan(a+b)=1/3 then tan2a=?

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if tan(a-b)=1/4, tan(a+b)=1/3 then tan2a=?

Oct 14, 2014

#2
+95369
+10

$$\\atan(\frac{1}{4})=A-B\qquad (1)\\\\ atan(\frac{1}{3})=A+B\qquad(2)\\\\ (1)+(2)\\\\ atan(\frac{1}{4})+atan(\frac{1}{3})=2A\\\\ tan[atan(\frac{1}{4})+atan(\frac{1}{3})]=tan[2A]\\\\$$

$$Now I am going to use the identity\\\\ \boxed{tan(\theta_1+\theta_2)=\frac{tan(\theta_1)+tan(\theta_2)}{1-tan(\theta_1)tan(\theta_2)}}\\\\\\ tan(2A)\\\\ =tan[atan(\frac{1}{4})+atan(\frac{1}{3})]\\\\ =\frac{\frac{1}{4}+\frac{1}{3}}{1-\frac{1}{4}\times\frac{1}{3}}\\\\ =\frac{\frac{7}{12}}{1-\frac{1}{12}}\\\\ =\frac{7}{12}\div \frac{11}{12}\\\\ =\frac{7}{12}\times \frac{12}{11}\\\\ =\frac{7}{11}\\\\ This answer is exact$$$. Oct 14, 2014 ### 4+0 Answers #1 +94619 +5 Using the tangent inverse tan-1(1/4) = (a - b) = about 14.04° ...... and tan-1(1/3) = (a +b) = about 18.43° So (a -b) + (a+b) = (14.04 + 18.43)° → 2a = about 32.47° So tan(2a) = tan(32.47°) = about .636363= 7/11 Oct 14, 2014 #2 +95369 +10 Best Answer $$\\atan(\frac{1}{4})=A-B\qquad (1)\\\\ atan(\frac{1}{3})=A+B\qquad(2)\\\\ (1)+(2)\\\\ atan(\frac{1}{4})+atan(\frac{1}{3})=2A\\\\ tan[atan(\frac{1}{4})+atan(\frac{1}{3})]=tan[2A]\\\\$$ $$Now I am going to use the identity\\\\ \boxed{tan(\theta_1+\theta_2)=\frac{tan(\theta_1)+tan(\theta_2)}{1-tan(\theta_1)tan(\theta_2)}}\\\\\\ tan(2A)\\\\ =tan[atan(\frac{1}{4})+atan(\frac{1}{3})]\\\\ =\frac{\frac{1}{4}+\frac{1}{3}}{1-\frac{1}{4}\times\frac{1}{3}}\\\\ =\frac{\frac{7}{12}}{1-\frac{1}{12}}\\\\ =\frac{7}{12}\div \frac{11}{12}\\\\ =\frac{7}{12}\times \frac{12}{11}\\\\ =\frac{7}{11}\\\\ This answer is exact$$$

Melody Oct 14, 2014
#3
+95369
+5

Thanks Chris

Oct 14, 2014
#4
+94619
+5

I did mine "on the fly"....Melody's approach is actually "better"

Oct 14, 2014