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Question Number 204083 by Tawa11 last updated on 05/Feb/24

Given that tan(A + B) = 1 and tan(A - B) = 1/7 find tan A and tan B
	Answered by a.lgnaoui last updated on 06/Feb/24
	$posons t1=tan A t2=tan B { ((((t1+t2)/(1−t1t2))= 1)),((((t1−t2)/(1+t1t2))=(1/7))) :} t1+t2=1−t1t2 t1−t2=(1/7)+(1/7)t1t2 2t1=((8−6t1t2)/7) t1=(4/(3t2+7)) t2+(4/(3t2+7))=1−((4t2)/(3t2+7)) 3(t2)^2 +8t2−3=0 △′=5^2 { ((t2=(1/6))),((t2=((−3)/2))) :} a) t2=(1/6) t1=(8/(15)) b)t2=((−3)/2) t1=−(8/(23)) tan A=(8/(15)) tan B=(1/6) tan A= (8/(23)) tan B=((−3)/2)$
	$posons t 1 = \tan A t 2 = \tan B$ ${\begin{cases} \frac{t 1 + t 2}{1 - t 1 t 2} = 1 \\ \frac{t 1 - t 2}{1 + t 1 t 2} = \frac{1}{7} \end{cases}$ $t 1 + t 2 = 1 - t 1 t 2$ $t 1 - t 2 = \frac{1}{7} + \frac{1}{7} t 1 t 2$ $2 t 1 = \frac{8 - 6 t 1 t 2}{7} t 1 = \frac{4}{3 t 2 + 7}$ $t 2 + \frac{4}{3 t 2 + 7} = 1 - \frac{4 t 2}{3 t 2 + 7}$ $3 {(t 2)}^{2} + 8 t 2 - 3 = 0$ $△^{'} = 5^{2} {\begin{cases} t 2 = \frac{1}{6} \\ t 2 = \frac{- 3}{2} \end{cases}$ $a) t 2 = \frac{1}{6} t 1 = \frac{8}{15}$ $b) t 2 = \frac{- 3}{2} t 1 = - \frac{8}{23}$ $\tan A = \frac{8}{15} \tan B = \frac{1}{6}$ $\tan A = \frac{8}{23} \tan B = \frac{- 3}{2}$
	Commented by Frix last updated on 06/Feb/24

	$Method ok but calculation error :$ $3 t_{2}^{2} + 8 t_{2} - 3 = 0$ $\Rightarrow t_{2} = - 3 \lor t_{2} = \frac{1}{3}$
	Commented by Tawa11 last updated on 06/Feb/24

	$I appreciate sir .$
	Commented by a.lgnaoui last updated on 07/Feb/24

	$thank you$
	Answered by cortano12 last updated on 06/Feb/24

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