Question

It tanA+ tantB + tanA x tanB = 1
then A+B=?
​

Answer 1

AnswEr :

Given that,

$\sf \: tan \: A + tan \: B + tan \: A.tan \: B = 1$

To finD :

The value of A + B.

Now,

$\implies \sf \: tan \: A + tan \: B = 1 - tan \: A.tan \: B$

Dividing by 1 - tan(A).tan(B) on both sides,

$\implies \sf \: \dfrac{tan \: A+ tan \: B}{1 - tan \: A.tan \: B}= \cancel{ \dfrac{ 1 - tan \: A.tan \: B}{1 - tan \: A.tan \: B}}$

$\implies \sf \: \dfrac{tan \: A+ tan \: B}{1 - tan \: A.tan \: B}= 1$

We know that,

$\star \: \boxed{ \boxed{ \sf tan(x + y) = \dfrac{tan \: x + tan \: y}{1 - tan \: x.tan \: y} }}$

Thus,

$\implies \sf \: tan(A + B) = 1 \\ \\ \implies \large{ \boxed{ \boxed{ \sf A+ B = 45}}}$

The value of A + B is 45°.

Answer 2

Question:

It tanA+ tanB + (tanA × tanB) = 1, then A+B=?

To find:

To find the value of A + B

Answer:

$The \: value \: of \: \underline\bold{A+B \: is \: 45°}$

Step-by-step explanation:

$tanA + tanB + (tanA \times tanB) = 1$

$\longrightarrow tanA + tanB = 1 - tanA \times tanB$

Dividing by 1 - tanA × tanB

$\longrightarrow \frac{tanA + tanB}{1 - tanA \times tanB} = \frac{1 - tanA \times tanB}{1 - tanA \times tanB}$

$\longrightarrow \frac{tanA + tanB}{1 - tanA \times tanB} = 1$

Therefore,

$\boxed{tan(x + y) = \frac{tanx + tany}{1 - tanx \times tany}}$

Now,

$\longrightarrow tan(A + B)= 1$

We know that,

$\boxed{tan 45° = 1}$

$\longrightarrow tan(A + B)= 1$

$\boxed{ A + B = 45° }$

Thus the value of A + B = 45°