A+B=(π/4) find the (1+tanA)(1+tanB)=? with explanotory solution


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Question Number 186590 by mustafazaheen last updated on 06/Feb/23

$A + B = \frac{π}{4} f i n d t h e (1 + \tan A) (1 + \tan B) = ?$ $w i t h e x p l a n o t o r y s o l u t i o n$
	Answered by mr W last updated on 06/Feb/23

	$\tan (A + B) = \tan \frac{π}{4} = 1$ $\frac{\tan A + \tan B}{1 - \tan A \tan B} = 1$ $\tan A + \tan B = 1 - \tan A \tan B$ $1 + \tan A + \tan B + \tan A \tan B = 2$ $1 + \tan A + (1 + \tan A) \tan B = 2$ $(1 + \tan A) (1 + \tan B) = 2 ✓$
	Answered by Frix last updated on 06/Feb/23

	$1 + \tan B = 1 + \frac{\sin B}{\cos B} = 1 + \frac{\sin (\frac{π}{4} - A)}{\cos (\frac{π}{4} - A)} =$ $= 1 + \frac{\frac{\cos A - \sin A}{\sqrt{2}}}{\frac{\cos A + \sin A}{\sqrt{2}}} = 1 + \frac{\cos A - \sin A}{\cos A + \sin A} =$ $= \frac{2 \cos x}{\cos x + \sin x} = \frac{2}{1 + \tan A}$ $\Rightarrow answer is 2$
	Answered by a.lgnaoui last updated on 06/Feb/23

	$(1 + \tan A) (1 + \tan B) = P$ $P = 1 + (\tan A + \tan B) + \tan A \times \tan B$ $w e k n o w t h a t \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \times \tan B}$ $\tan A + \tan B = \tan (A + B) [1 - \tan A \times \tan B]$ $= 1 - \tan A \times \tan B$ $P = 1 + 1 - \tan A \times \tan B + \tan A \times \tan B = 2$ $d o n c$ $(1 + \tan A) (1 + \tan B) = 2$
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