A-B-pi-4-find-the-1-tanA-1-tanB-with-explanotory-solution-

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) = \boldsymbol P

\boldsymbol 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

\boldsymbol 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