Question

if A-B = 45 degree , show that (1+ tanA ) * (1 + tanB ) = 2tanA .

Answer 1

\[\large{A-B = \cfrac{\pi}{4}}\]

\[\large{\implies \tan(A-B) = \tan \cfrac{\pi}{4} = 1}\]

\[\large{\implies \cfrac{\tan A - \tan B}{1+\tan A \tan B} = 1}\]

\[\large{\implies \tan A - \tan B = 1 + \tan A \tan B}\]

\[\large{\implies \tan A = 1+\tan B + \tan A ~ \tan B}\tag{1}\]

Answer 2

Now:

\[\large{(1+\tan A)(1+\tan B) = 1+\tan A + \tan B + \tan A ~\tan B}\]

\[\large{= \tan A + (1+\tan B + \tan A ~ \tan B)}\tag{2}\]

Answer 3

Now from equation (1) use the value of :

1+ tan B + tan A * tan B = tan A

Answer 4

Put this value in equation (2)

Answer 5

you would get:

tan A + (tan A)

= 2 tan A

Answer 6

\[\large{\color{red}{\text{Hence Proved}}}\]

Answer 7

@miniash11 Any doubt ? :)

Answer 8

umm...yes... how did u get the equation 2 ? by adding 1 on both sides ?

Answer 9

No no I did nothing like that

I just collected `1 + tan B + tan A tan B` together inside `( )` brackets and left the `tan A` outside the brackets.

Answer 10

The only formula I used here is:

\[\large{\boxed{\tan (x-y) = \cfrac{\tan x - \tan y}{1 + \tan x \tan y}}}\] in:

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5

\[\large{A-B = \cfrac{\pi}{4}}\]

\[\large{\implies \color{red}{\tan(A-B)} = \tan \cfrac{\pi}{4} = 1}\]

\[\large{\implies \color{red}{\cfrac{\tan A - \tan B}{1+\tan A \tan B}} = 1}\]

\(\color{blue}{\text{End of Quote}}\)

Answer 11

Any doubt ? :)

Answer 12

yes sir, i did understand that but the only point im stuck at is
(1+tanA)(1+tanB)=1+tanA+tanB+tanA tanB -> this line


=tanA+(1+tanB+tanA tanB)(2)

Answer 13

Please don't call me sir. I am just a student :)

Call me Vishwesh or Vishu

Answer 14

Okay lets see if I can help you out in that

Answer 15

first of all expand (1+tan A)(1 + tan B)

What will you get ?

Answer 16

okay vishu

Answer 17

:D Hi @dg2

How are you ?

Answer 18

1 + tanB + tanA + 1 + tanAtanB ?

Answer 19

Why that 1 again ?