Question

Hola brainlians ! Todays Question ..

If A+B = 135° then,

$\sf\dfrac{tanA .tanB }{(1+tanA)(1+tanB)}$ = ?

____________________________

Hint :- answer is 1/2




No spamming..Pls help me​

Answer 1

Answer:

here it is .. but I got 1 as a answer . if you guys found any mistakes with mine pls comment. thanks

Answer 2

Answer:

Given :

$\leadsto \sf A\: +\: B\: = 135^{\circ}$

To Find :-

$\leadsto \sf What\: is\: the\: value\: of\: \sf \dfrac{tanA . tanB}{(1 + tanA)(1 + tanB)}$

Solution :-

$\implies \sf A + B =\: 135^{\circ}$

Then, we can write as,

$\implies \sf A + B =\: 180^{\circ} - 45^{\circ}$

$\implies \sf\bold{\purple{B =\: 180^{\circ} - (A + 45^{\circ})}}$

Now,

$\implies \sf \dfrac{tanA . tanB}{(1 + tanA)(1 + tanB)}$

We can simply put the value of B,

$\implies \sf \dfrac{tanA . tan\bigg(180^{\circ} - (A + 45^{\circ})\bigg)}{(1 + tanA)(1 + tan\bigg(180^{\circ} - (A + 45^{\circ})\bigg)}$

$\implies \sf \dfrac{- tanA . tan(A + 45^{\circ})}{(1 + tanA)\bigg(1 + tan(A + 45^{\circ})\bigg)}$

$\implies \sf \dfrac{- tanA . \bigg(\dfrac{\cancel{1 + tanA}}{1 - tanA}\bigg)}{(\cancel{1 + tanA})\bigg(1 - \dfrac{1 + tanA}{1 - tanA}\bigg)}$

$\implies \sf \dfrac{- tanA . \dfrac{1}{\cancel{1 - tanA}}}{\bigg(\dfrac{\cancel{1} - tanA \cancel{-} 1 - tanA}{\cancel{1 - tanA}}\bigg)}$

$\implies \sf \dfrac{\cancel{-} \cancel{tanA}}{\cancel{-} 2\: \cancel{tanA}}$

$\implies \sf\bold{\red{\dfrac{1}{2}}}$

$\sf\boxed{\bold{\green{\therefore The\: value\: of\: \dfrac{tanA . tanB}{(1 + tanA)(1 + tanB)}\: is \dfrac{1}{2}\: .}}}$